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Answer:

Hindi ko Po alam sorry po

Answer:

no equation

Explanation:

Explanation:

the answer is in the image above

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) =

Answer:

∵

Explanation:

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.

Answer:

Wavelength

a. Increases as energy decreases

b. Often measured in units such as nanometers (nm)

f. A measure of distance

Frequency

c. Often measured in units such as Hertz (Hz)

d. Increases as energy increases

e. A measure of cycles per unit time

Explanation:

Wavelength increases when the energy decreases whereas frequency increases as energy increases. Wavelength is measured in nanometers (nm) whereas frequency is measured in hertz. Frequency refers to the number of cycles in a unit of time while on the other hand, the distance between successive crests of a wave, especially in a sound wave or electromagnetic wave is known as wavelength.

Answer:

iv. N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺

Explanation:

Isoelectronic elements are those that have the same number of electrons. So, if at least 2 elements differ in their number of electrons, the series is not of isoelectronic elements.

To know the number of electrons we will consider the atomic number and add electrons if it is an anion and subtract electrons it is a cation.

Identify the isoelectronic elements.

Answer:

a)the reaction will proceed to the left, the mass of CaO will decrease.

b)the reaction is in equilibrium, mass of CaO will not change.

c)the reaction is in equilibrium, mass of CaO will not change.

d) the reaction will proceed to the right, the mass of CaO will increase.

Explanation:

Step 1: Data given

At 900°C, Kp = 1.04  ( Kp is only for compounds that are in gaseous state)

Volume of the chamber = 50.0L

Temperature is raised to 900 °C

Step 2: The reaction

CaCO31(s) ⇔ CaO (s) + CO2 (g)

Kp for CO2 = 1.04

Step 3: First mix :  655g CaCO3, 95.0g CaO, PCO2 = 2.55 atm

We only consider CO2 here

pCO2 > Kp  this means the reaction will proceed to the left, the mass of CaO will decrease.

Step 4: 780g CaCO3, 1.00g CaO, PCO2 = 1.04 atm.

We only consider CO2 here

pCO2 = Kp  this means the reaction is in equilibrium, mass of CaO will not change.

Step 5: 0.14g CaCO3, 5000g CaO, PCO2 = 1.04 atm

We only consider CO2 here

pCO2 = Kp  this means the reaction is in equilibrium, mass of CaO will not change.

Step 6: 715g CaCO3, 813g CaO, PCO2 = 0.211 atm

We only consider CO2 here

pCO2 < Kp  this means the reaction will proceed to the right, the mass of CaO will increase.

Answer: in 1975, the United States passed the Metric Conversion Act. The legislation was meant to slowly transition its units of measurement from feet and pounds to meters and kilograms, bringing the US up to speed with the rest of the world. There was only one issue: the law was completely voluntary. Of course, that meant it pretty much never took off

Explanation:

they passed the metric act

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.

Answer:

b

Explanation:

bc why not

Answer:

Which type of chemistry studies the chemical reactions that occur in the human body? .

biochemistry

The correct answer is biochemistry.

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Answer:

3Ca²⁺(aq) +  2PO4³⁻(aq) → Ca3(PO4)2(s)

Explanation:

The calcium of CaCl2 reacts with the phosphate ion od Cs3PO4 to produce the insoluble salt Ca3(PO4)2 and CsCl. The unbalanced reaction is:

CaCl2(aq) + Cs3PO4(aq) → Ca3(PO4)2(s) + CsCl(aq)

To balance the calciums:

3CaCl2(aq) + Cs3PO4(aq) → Ca3(PO4)2(s) + CsCl(aq)

The chlorides:

3CaCl2(aq) + Cs3PO4(aq) → Ca3(PO4)2(s) + 6CsCl(aq)

And the Cs:

3CaCl2(aq) + 2Cs3PO4(aq) → Ca3(PO4)2(s) + 6CsCl(aq)

This is the balanced reaction, the ionic equation is:

3Ca²⁺(aq) + 6Cl⁻(aq) + 6Cs⁺(aq) + 2PO4³⁻(aq) → Ca3(PO4)2(s) + 6Cs⁺(aq) + 6Cl⁻(aq)

Subtracting the ions that don't react:

This is the net ionic equation

The balanced, net ionic equation for the precipitation reaction of CaCl₂ and Cs₃PO₄ in an aqueous solution can be written as follows:

3Ca²⁺(aq) + 2PO₄³⁻(aq) -> Ca₃(PO₄)₂(s)

A balanced equation is a chemical equation in which the number of atoms of each element on both sides of the equation is equal. It follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

In a balanced equation, the coefficients in front of each compound or molecule indicate the number of moles or molecules involved in the reaction. By adjusting these coefficients, the equation is balanced in terms of both mass and charge.

In this equation, Ca²⁺ ions from calcium chloride (CaCl₂) combine with PO₄³⁻ ions from cesium phosphate (Cs₃PO₄) to form solid calcium phosphate (Ca₃(PO₄)₂). The net ionic equation represents only the species directly involved in the reaction and excludes spectator ions, which do not participate in the formation of the precipitate. The physical states are denoted as (aq) for aqueous and (s) for solid.

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Answer:

Q = 6019J

Explanation:

Using the formula as follows:

Q = m × c × ∆T

Where;

c = specific heat capacity

Q = amount of heat (J)

m = mass of substance

∆T = change in temperature (°C)

According to the information given in this question,

m = 80g

c = 4.18 J/g°C

Initial temperature = 25°C

Final temperature = 43°C

∆T = 43°C - 25°C = 18°C

Q = 80 × 4.18 × 18

Q = 6019.2J

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)

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]

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

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]

Answer:

For the neutralization reaction between pyridine and propanoic acid, draw curved arrows to indicate the direction of electron flow.

Draw curved arrows to show the movement of electrons in this step of the mechanism.

Explanation:

According to Bronsted acid-base theory, an acid is a substance which is a proton donor.

Base is the proton acceptor.

In the given example, acid is propanoic acid and it loses the proton.

Pyridine is the base and it accepts the proton from propanoic acid.

The entire reaction is shown below:

Answer: The value of [tex][H_{3}O^{+}][/tex] is [tex]6.025 \times 10^{-12}[/tex] M and [tex][OH^{-}][/tex] is [tex]1 \times 10^{-14}[/tex].

Explanation:

pH is the negative logarithm of concentration of hydrogen ion.

It is given that pH is 11.22. So, the value of concentration of hydrogen ions is calculated as follows.

[tex]pH = - log [H^{+}]\\11.22 = - log [H^{+}]\\conc. H^{+} = 6.025 \times 10^{-12}[/tex] M

Let the value [tex]6.025 \times 10^{-12}[/tex] is considered as equal to 0. Hence, the relation between pH and pOH value is as follows.

pH + pOH = 14

0 + pOH = 14

pOH = 14

Now, pOH is the negative logarithm of concentration of hydroxide ions.

Hence, [tex][OH^{-}][/tex] is calculated as follows.

[tex]pOH = - log [OH^{-}]\\14 = - log [OH^{-}]\\conc. OH^{-} = 1 \times 10^{-14} M[/tex]

Thus, we can conclude that the value of [tex][H_{3}O^{+}][/tex] is [tex]6.025 \times 10^{-12}[/tex] M and [tex][OH^{-}][/tex] is [tex]1 \times 10^{-14}[/tex].

Answer:

3.9g/cm3

Explanation:

Density ( d)=?

Mass(m)=27.3g

Volume (v)=7.0cm3

D=m÷v

D=27.3g÷7.0cm3

D=3.9g/cm3

Answer:

The system is not at equilibrium and the reaction will proceed to the left.

Explanation:

Step 1: Write the balanced equation

H₂(g) + CO₂(g) ⇄ CO(g) + H₂O(g)

Step 2: Calculate the reaction quotient (Q)

The reaction is calculated in the same way as the equilibrium constant (Kc) but it uses the concentrations at any time.

Q = [CO] × [H₂O] / [H₂] × [CO₂]

Q = 0.610 × 0.695 / 0.425 × 0.500 = 2.00

Since Q ≠ Kc, the reaction is not at equilibrium.

Since Q > Kc, the reaction will proceed to the left.

Answer:

Im sorry i cant write a whole essay right now but these sourcea could help.

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Answer:

have you done it yet?

Explanation:

..

Answer:

It is A, The daughter product has the wrong atomic number

Explanation:

Have trust, please i beg of you.

:) thank me later, love ya

The daughter's product has the wrong atomic number. Hence, option A is correct.

Nuclear equations represent the reactants and products in radioactive decay, nuclear fission, or nuclear fusion.

Beta decay - A common mode of radioactive decay in which a nucleus emits beta particles. The daughter nucleus will have a higher atomic number than the original nucleus.

Hence, the daughter's product has the wrong atomic number.

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Actual yield over theoretical yield, then multiply by 100

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].

Answer:

D

Explanation:

D

The mode of decay for each of the following unstable nuclides are (a) ⁵⁷Ti is beta decay (b) ⁵⁹Zn is Positron emission (c) ¹²³Ba is Positron emission (d) ⁶⁰Cr is beta decay.

Now lets predict the mode of decay for each of the following:

(a) Ti - 57

⁵⁷Ti₂₂ → ⁵⁷V₂₃  + ⁰e₋₁

Here A is unchanged and Z is increases by 1 so the mode of decay is beta decay.

(b) Zn - 59

⁵⁹Zn₃₀ → ⁵⁹Cu₂₉ + ⁰e₊₁

Here A is unchanged and Z is decreases by 1 so the mode of decay is Positron emission.

(c) Ba - 123

¹²³Ba₅₆ → ¹²³Cs₅₅ + ⁰e₊₁

Here A is unchanged and Z is decreases by 1 so the mode of decay is Positron emission.

(d) Cr -60

⁶⁰Cr₂₄ → ⁶⁰Mn₂₅ + ⁰e₋₁

Here A is unchanged and Z is increases by 1 so the mode of decay is beta decay.

Thus, from above conclusion  we can say that (a) ⁵⁷Ti is beta decay (b) ⁵⁹Zn is Positron emission (c) ¹²³Ba is Positron emission (d) ⁶⁰Cr is beta decay.

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Answer:

B

Explanation:

the concentration will be the same disabled

Answer:

[tex]{ \tt{22.4 \: dm {}^{3} \: contains \: 1 \: mole \: of \: hydrogen }} \\ { \tt{0.15 \: {dm}^{3} \: will \: contain \: ( \frac{0.15}{22.4} \times 1) \: moles }} \\ = 0.0067 \: moles \\ { \bf{mole \: ratio = 1 :1 }} \\ { \tt{1 \: mole \: weighs \: 65.38 \: g}} \\ { \tt{0.0067 \: moles \: weighs \: (0.0067 \times 65.38) \: g}} \\ = 0.44 \: g[/tex]

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