Match all the terms that apply to each element. Some terms may apply to more than one element.

a. Main group element
b. halogen
c. noble gas
d. inner transition element
e. alkaline earth
f. Non-metal
g. Alkali metal
h. Metalloid
i. Metal

1. Titanium
2. Gold
3. Argon
4. Thallium

Answers

Answer 1

Answer:

Titanium - Metal

Gold - Metal

Argon - noble gas

Thallium - Metal

Explanation:

A metal is a species that easily looses electrons to yield a positive ion. Metals are found in the left hand side of the periodic table.

Thallium is a groups 13 metal, gold is a group 11 metal while titanium is a 4 metal.

However, argon is a noble gas hence it belongs to groups 18 of the periodic table.

Answer 2

Titanium -metal

gold - metal,

Argon - noble gas

Thallium -metal, transition metal

Explanation:

In modern periodic table :

Metals are listed from the middle to left-hand side of the periodic tableNonmetals are listed on the upper right-hand side of the periodic tableMetalloids divide the periodic table in a zig-zag line with nonmetals on the right and metals on the left.The alkali metals are listed under group 1The alkaline earth metals are listed under group 2The noble gases are listed under group 18The transition metals are listed from group 3 to group 12.The halogens are listed under group 17The inner transition elements (f-block) are listed at the bottom separately.

So, from this, we can match the given elements as:

Titanium -metal

gold - metal,

Argon - noble gas

Thallium -metal. transition metal

Learn more about the periodic table:

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Match All The Terms That Apply To Each Element. Some Terms May Apply To More Than One Element. A. Main

Related Questions

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Answers

Answer:

Approximately [tex]1.30 \times 10^{-2}[/tex], assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let [tex]\rm HA[/tex] denote this acid.

[tex]\rm HA \rightleftharpoons H^{+} + A^{-}[/tex].

Initial concentration of [tex]\rm HA[/tex] without any dissociation:

[tex][{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}[/tex].

After [tex]12.5\%[/tex] of that was dissociated, the concentration of both [tex]\rm H^{+}[/tex] and [tex]\rm A^{-}[/tex] (conjugate base of this acid) would become:

[tex]12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}[/tex].

Concentration of [tex]\rm HA[/tex] in the solution after dissociation:

[tex](1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}[/tex].

Let [tex][{\rm HA}][/tex], [tex][{\rm H}^{+}][/tex], and [tex][{\rm A}^{-}][/tex] denote the concentration (in [tex]\rm mol \cdot L^{-1}[/tex] or [tex]\rm M[/tex]) of the corresponding species at equilibrium. Calculate the acid dissociation constant [tex]K_{\rm a}[/tex] for [tex]\rm HA[/tex], under the assumption that this acid is monoprotic:

[tex]\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}[/tex].

The pH of a 0.1 M solution of an unknown weak acid is 3.7. What is the pKa of this acid?

Answers

Explanation:

Knowing the pH, you know the concentration of protons:

−log[H+]=pH=3.7

[H+]=10−3.7 M

Now, since the weak (monoprotic) acid dissociates into its conjugate base and a proton, the mols of protons are equimolar with the mols of conjugate base---the protons came FROM the weak acid, so the conjugate base that forms must be equimolar with the protons given out to the solvent.

HA⇌A−+H+

Hence, [A−]=[H+] in the same solution volume. Using the equilibrium constant expression, we get:

Ka=[H+]2eq[HA]eq

Don't forget that the HA form of HA had given away protons, so the mols of protons given away to generate A− is subtracted from the mols of (protons in) HA.

=[H+]2eq[HA]i−[H+]eq

=(10−3.7M)20.02M−10−3.7M

Ka=2.0105×10−6 M

in a reaction 2Fe+Cl2→2FeCl2 ​

Answers

Answer:

no equation

Explanation:

A 8.29g sample of calcium sulfide was decomposed into its constituent elements, producing 4.61g of calcium and 3.68g of sulfur. Which of the statements are consistent with the law of constant composition (definite proportions)?

a. Every sample of calcium sulfide will have 44.4% mass of calcium.
b. Every sample of calcium sulfide will have 2.86 g of calcium.
c. The mass ratio of Ca to S in every sample of calcium sulfide is 1.25.
d. The ratio of calcium to sulfur will vary based on how the sample was prepared.
e. The mass percentage of calcium plus the mass percentage of sulfur in every sample of calcium sulfide equals 100%.

Answers

Answer:

d,e

Explanation:

If 7.90 mol of C5H12 reacts with excess O2, how many moles of CO2 will be produced by the following combustion reaction?

C5H12+8O2=6H2O+5CO2

Answers

Answer:

moles CO₂ produced from combustion of 7.9 mole of C₅H₁₂ = 39.5 moles CO₂

Explanation:

              C₅H₁₂      +   8O₂      =>      6H₂O      +      5CO₂

Given:   7.9moles    excess            _____            ? moles

             

From Equation, 1 mole C₅H₁₂ =============>  5 moles CO₂

Given            7.9 mole C₅H₁₂  =============>         X

Solving for 'X' using ratio and proportion ...

1mole C₅H₁₂ / 7.9mole C₅H₁₂ =  5mole CO₂ / X

=> X = 5 mole CO₂ x  7.9mole C₅H₁₂ / 1mole C₅H₁₂  =  39.5 moles CO₂

≅ 40 moles CO₂ (2 sig. figs.)

What is the observation of heating of iodine crystals

Answers

Answer:

On heating, the van der Waals dispersion forces existing then will easily break as it has a low boiling point and sublimates into gas. On heating iodine in the test tube, iodine evolves as violet fuming gas.

Explanation:

How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?

Answers

Answer:

0.375mL of solution of nalorphine must be injected

Explanation:

A solution of 0.40% (w/v) contains 0.40g of solute (In this case, nalorphine), in 100mL of solution. To obtain 1.5mg of nalorphine = 1.5x10⁻³g of nalorphine are needed:

1.5x10⁻³g * (100mL / 0.40g) =

0.375mL of solution of nalorphine must be injected

why is platinum metal preferred to other metals for the flame test​

Answers

Answer:

Platinum is especially good for this because it is unreactive, and does not produce a color in the flame which will mask the presence of other metals.

Hope this answer is right!

Answer:

Hey mate, here is your answer

1. Platinum doesn't impart any color to the flame.

2. It is not oxidised under the high temperature of the flame from a bunsen burner.

3. It is almost chemically inert. Even at high temperatures, it remains unattacked by free radicals / acid radicals.

Therefore, platinum wire is crucial for a flame test. Also, a platinum wire should be thoroughly cleaned before using it for a new flame test.

A platinum wire is cleaned by dipping it into concentrated HNO3 and then placing it in the non luminous part of the bunsen flame. Otherwise, the perviously tested radicals will impart color to the flame, which may cause confusion.

Explanation:

Hope it helps you

Balancea las siguientes ecuaciones por el método de redox. Determina el
agente oxidante, el agente reductor, la sustancias oxidada y la sustancia
reducida para cada caso.
a) C + H2SO4
CO2 + SO2 + H2O
I
b) sh + HNO3
Sb2O5 +NO + H2O
c) Bi(OH)3 +K SnO2
Bi + K Sn03
+ H2O
d) H2S + (Cr2O7)2- + H1+
H2O + S + Cr3+

Answers

Answer:

Traduzca al ingles para que pueda responder a su pregunta. Te dare la reapuesta en los comentarios

D. Write the name of the branched alkane next to the drawing of the molecule. (2 points)

Answers

Answer: hello your question lacks some data attached below is the missing data

answer :

a) 3-methyl heptane

b) 2-methyl pentane

c) 2-methyl heptane

d) 2-methyl hexane

e) 3-methyl hexane

Explanation:

we will select the longest carbon chain as the branched alkane and name it

a) 3-methyl heptane ( first diagram )

b) 2-methyl pentane ( second diagram )

c) 2-methyl heptane ( third diagram )

d) 2-methyl hexane ( fourth diagram )

e) 3-methyl hexane ( fifth diagram )

Note : sixth diagram = first diagram

what two parts are needed to make a neutral atom of neon

Answers

Answer:

it needs two electrons in the first and eight to fill the second.

Explanation:

Because neon has two atomic shells, it needs two electrons in the first and eight to fill the second. Neon has a total of ten electrons which means two filled shells.

2) If the density of chloroform is 1.48 g/mL, what is the volume of 541 g of chloroform?

Answers

Answer:

V = 365.54 mL

Explanation:

Given that,

The density of chloroform, d = 1.48 g/mL

The mass of chloroform, m = 541 g

We need to find the volume of chloroform.

We know that,

Density = mass/volume

So,

[tex]V=\dfrac{m}{d}\\\\V=\dfrac{541\ g}{1.48\ g/mL}\\\\=365.54\ mL[/tex]

So, the volume of chloroform is 365.54 mL.

Question 3 (2 points)
Using the following equation how many grams of water you would get from 846 g of
glucose:
C6H12O6 + 602 + 6CO2 + 6H20 + energy
Your Answer:
Answer
units g)

Answers

Answer: The mass of water produced is 507.6 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of glucose = 846 g

Molar mass of glucose = 180 g/mol

Plugging values in equation 1:

[tex]\text{Moles of glucose}=\frac{846g}{180g/mol}=4.7 mol[/tex]

The given chemical equation follows:

[tex]C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O+energy[/tex]

By the stoichiometry of the reaction:

If 1 mole of glucose produces 6 moles of water

So, 4.7 moles of glucose will produce = [tex]\frac{6}{1}\times 4.7=28.2mol[/tex] of water

Molar mass of water = 18 g/mol

Plugging values in equation 1:

[tex]\text{Mass of water}=(28.2mol\times 18g/mol)=507.6g[/tex]

Hence, the mass of water produced is 507.6 g

Jax designs an experiment to determine how the amount of sodium chloride affects the boiling point of water. He adds 1 g, 5 g, and 10 g of sodium chloride to three different beakers, each containing 100 mL of water. There is a fourth beaker that contains 100 mL water without any sodium chloride. He heats each of the samples on a Bunsen burner and measures the boiling point with the same thermometer. Which of the following is/are the control(s) in the experiment? (Choose all that apply)

Answers

Answer:

Amount of water

The thermometer

Explanation

In an experiment, there is always a dependent variable and an independent variable. The independent variable is manipulated and its effect on the dependent variable is observed.

The control is that factor in the experiment that must remain constant so that effect of the independent variable on the dependent variable can be observed.

In this case, the independent variable is the amount of sodium chloride while the dependent variable is the temperature at which the solution boils.

The controls must be the amount of water which must be held constant and the same thermometer used to measure the temperature so that the effect of the amount of sodium chloride on the temperature of the solution can be studied.

The following are the controls in the experiment:

The beaker with 100 mL of water without any sodium chloride.

The temperature of the Bunsen burner.

The type of thermometer used.

The control(s) in the experiment are the beaker with 100 mL of water without any sodium chloride. This beaker is used to compare the boiling points of the other beakers, which have different amounts of sodium chloride added.

The control beaker ensures that any differences in boiling point are due to the amount of sodium chloride added, and not to other factors, such as the temperature of the Bunsen burner or the type of thermometer used.

The other factors that could affect the boiling point of water, such as the humidity of the air or the altitude, are kept constant in the experiment.

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Groups on the periodic table also correspond with the number of ?

Answers

The question is incomplete, the complete question is;

Groups of the periodic table correspond to elements with a. the same color b. the same atomic number c. similar chemical properties d. similar numbers of neutrons

Answer:

similar chemical properties

Explanation:

In the periodic classification of elements, elements are divided into groups and periods. Elements in the same group of the periodic table have the same number of outermost electrons and share very similar chemical properties.

Elements in the same period have the same number of shells and the same maximum energy level of the outermost electron. Chemical properties carry markedly across a period.

Liquid hexane
(CH,(CH), CH) will react with gaseous oxygen (0) to produce gaseous carbon dioxide (CO2) and gaseous water (1,0). Suppose 1.72 g
of hexane is mixed with 8.0 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the
correct number of significant digits.

Answers

Answer: The mass of [tex]H_2O[/tex] produced is 2.52 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

For hexane:

Given mass of hexane = 1.72 g

Molar mass of hexane = 86.18 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of hexane}=\frac{1.72g}{86.18g/mol}=0.020mol[/tex]

For oxygen gas:

Given mass of oxygen gas = 8.0 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of water}=\frac{8.0g}{32g/mol}=0.25mol[/tex]

The chemical equation for the combustion of hexane follows:

[tex]2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O[/tex]

By stoichiometry of the reaction:

If 2 moles of hexane reacts with 19 moles of oxygen gas

So, 0.020 moles of hexane will react with = [tex]\frac{19}{2}\times 0.020=0.19mol[/tex] of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hexane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of hexane produces 14 moles of [tex]H_2O[/tex]

So, 0.020 moles of hexane will produce = [tex]\frac{14}{2}\times 0.020=0.14mol[/tex] of [tex]H_2O[/tex]

We know, molar mass of [tex]H_2O[/tex] = 18 g/mol

Putting values in above equation, we get:

[tex]\text{Mass of }H_2O=(0.14mol\times 18g/mol)=2.52g[/tex]

Hence, the mass of [tex]H_2O[/tex] produced is 2.52 g

Which of the following reasons explains why if salt water is heated, the water turns into steam while the salt remains?

Water and salt have an equal boiling point.
Water has a lower boiling point than salt.
Salt has a lower boiling point than water.
Salt and water have an equal melting point.

Answers

If the salt water is heated, the water turns into steam and the salt remains because the water has a lower boiling point than the salt.

The  following points can be considered:

The boiling point is defined as the temperature at which the substance turns into the gaseous state from the liquid state.The boiling point of water is [tex]100^{o} C[/tex].The salt is a substance comprising two entities separated by the opposite charges with ionic interactions.The boiling point of a salt is higher than the boiling point of the water.

The process involved when salt water is heated:

The salt water mixture when heated, the water turns into steam at [tex]100^{o} C[/tex]But the salt remains until it reaches its boiling temperature. If the salt is soluble in water and is then heated, then there occurs an elevation in the boiling point of the substance, due to the presence of the salt.

Therefore, the answer is water has a lower boiling point than salt.

Learn more about salt:

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Question 1
Consider the following reaction:
Cl2(g) + 3 F2(g) —> 2 CIF3 (8)
How many moles of product will form if 0.115 moles of fluorine gas react?

Answers

Answer:

0.077 mole of ClF₃.

Explanation:

The balanced equation for the reaction is given below:

Cl₂ + 3F₂ —> 2ClF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of ClF₃.

Finally, we shall determine the number of mole of ClF₃ produced by the reaction of 0.115 mole of F₂. This can be obtained as follow:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of ClF₃.

Therefore, 0.115 mole of F₂ will react to to produce = (0.115 × 2)/3 = 0.077 mole of ClF₃.

Thus, 0.077 mole of ClF₃ was obtained from the reaction.

Calculate and compare the number of ATP molecules generated between a 24 carbon fatty acid and 4 glucose molecules.

Answers

Answer:

Idont no men sorry idont understand

Explanation:

Sorry

ACTUAL YIELD VS THEORETICAL YIELD?

Answers

Actual yield over theoretical yield, then multiply by 100

The main consequence of exposure to the chemicals at Love Canal was:
allergies.
hair loss.
birth defects.
upper respiratory disease.

Answers

The main consequence of exposure to the chemicals at Love Canal was:

allergies.

hair loss.

birth defects.☑️

upper respiratory disease.

The volume of a fixed mass of gas at 2 atm pressure is 20L.What will be its volume if the pressure is increased 4 times without changing the temperature.

Answers

Answer:

The correct answer is - 5L.

Explanation:

From the ideal gas equation -

pv=nRT

p = pressure

v =volume

Here nRT is constant so

P would be inversely proportioned to v

So, p1/p2 = v2/v1

Putting values:

2/4(2) = v2/20 (p2 = 4 times of P1)

2/8 = v2/20

V2 = 5

Thus, the correct volume at new pressure would be - 5 L.

 

If a light bulb has a voltage of 1.5 V and has 2.5A of current running through it, what is the resistance of the light bulb?

Answers

Answer:

0.6 ohm

Explanation:

voltage = 1.5 V

current = 2.5 A

resistence = ?

V = IR

1.5 = 2.5 * R

1.5 / 2.5 = r

0.6 = R

therefore resistence is 0.6 ohm

A series of measurements in the lab led to an experimental result of 32.9 mL, with a calculated standard deviation of 0.3 mL. What is the standard way to report this result?

Answers

Answer: The standard way to report this result is [tex]32.9\pm 0.3 mL[/tex]

Explanation:

The standard method of representing a result is:

[tex]\text{Calculated value }\pm \text{ Standard deviation}[/tex]

The reporting of a result is done in correct significant figures.

We are given:

Calculated value = 32.9 mL

Standard deviation = 0.3 mL

Rule of significant figures applied when numbers are added or subtracted:

The number having less number of significant figures after the decimal point will determine the number of significant figures in the final answer.

Number of significant figures after the decimal point = 1

Hence, the standard way to report this result is [tex]32.9\pm 0.3 mL[/tex]

Determine the molality and molarity of a sodium chloride solution prepared by adding 52.80 g of solid NaCl to 150.0 g of water. The density of the saline solution is 1.1972 g mL . Molality

Answers

Answer:

6.02 m5.334 M

Explanation:

First we convert 52.80 g of NaCl to moles, using its molar mass:

52.80 g ÷ 58.44 g/mol = 0.9035 mol

To determine the molality, we convert 150.0 g of water to kg:

150.0 g / 1000 = 0.150 kgmolality = 0.9035 mol / 0.150 kg =  6.02 m

As for the molarity, we determine the total mass of solution:

52.80 g + 150.0 g = 202.8 g

And calculate the volume, using the density:

202.8 g ÷ 1.1972 g/mL = 169.39 mL169.39 mL / 1000 = 0.16939 L

Finally we calculate the molarity:

0.9035 mol / 0.16939 L = 5.334 M

The strongest base that can exist in a solution in appreciable concentration is the conjugate base of the solvent.

a. True
b. False

Answers

Answer:

a. True

Explanation:

From the basic concepts of acids and bases, we know that when a base accepts a hydrogen ion (H⁺), it forms a conjugate acid which can accept again the H⁺ ion:

B⁻ + H₂O ⇆ BH + OH⁻

The stronger the base, the weaker the conjugate base. Thus, as more strength has a base, lesser strength will have the conjugate base (it will not be able to accept again the H⁺ ion). For example, when water (H₂O) loses its H⁺ , it forms the conjugate base OH⁻. So, OH⁻ is the stronger base that can exist in an aqueous solution.

H₂O ⇆ H⁺ + OH⁻

In fact, strong bases are hydroxides, such as NaOH or KOH.

The theoretical yield of zinc oxide in a reaction is 486 g. What is the percent
yield if 399 g is produced?
O A. 122%
O B. 4.93%
C. 82.1%
D. 29.6%

Answers

Answer:

the correct answer is c

Explanation:

becuase i had the same question

an element E forms a hydride EH3, which contains 90% of E by mass. what is the relative atomic mass ?​

Answers

Answer:

27 g/mol of E

Explanation:

Note that percentage by mass= mass of each element present. So, since there is 90% of E, there is 90g of E present. By implication, there are 10g of H corresponding to 10%H. Note that there is 100g of EH3

1 moles of E corresponds to 90 g of E

Mole ratio of E: H= 1:3

Thus

Number of moles of H = 10g/ 1g/ mol = 10 moles of H

Since E contains 1/3 the number of moles of H

Number of moles of E = 1/3 × 10 = 3.33 moles of E

Molar mass of E= mass of E/ number of moles of E

Since mass of E = 90 g

Molar mass of E = 90g/3.33 moles

Molar mass of E = 27 g/mol

According to the equation 2K(s) + CI2(g) 2kCI(s), potassium reacts with chlorine to form potassium chlorine. If 100 atoms of potassium react with chlorine gas, how many chlorine molecules will be needed to completely react?

Answers

Answer:

50 CI₂ molecules

Explanation:

2K(s) + CI₂(g) → 2KCI(s)

By looking at the stoichiometric coefficients, we can tell that if 2 atoms of potassium (K) react with chlorine gas (CI₂), 1 chlorine molecule would react.

With that in mind we can calculate how many CI₂ molecules would react with 100 K atoms:

100 atoms K *[tex]\frac{1Cl_2Molecule}{2KAtom}[/tex] = 50 CI₂ molecules

Which of the following amino acid residues would provide a side chain capable of increasing the hydrophobicity of a binding site?
A) aspartic acid
B) lysine
C) isoleucine
D) arginine
E) serine

Answers

Answer:

C) isoleucine

Explanation:

Isoleucine is among nine necessary amino acids in humans (found in dietary proteins). It has a variety of physiological activities, including aiding tissue repair, nitrogenous waste detoxification, immunological stimulation, and hormonal production promotion. When attached at a binding site, they are capable of providing a side chain thereby increasing the hydrophobicity at the binding site.

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