The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).

Answer:

371km

Explanation:

the lower the more fuel there is

Answer:

stratosphere

Explanation:

contains a high concentration of ozone in relation to other parts of the atmosphere, although still small in relation to other gases in the stratosphere.

Answer:

stratosphere

Explanation: Most atmospheric ozone is concentrated in a layer in the stratosphere, about 9 to 18 miles (15 to 30 km) above the Earth's surface. Ozone is a molecule that contains three oxygen atoms.

Answer:

We are given that there is 95% ethanol by mass in rectified spirit

so, we can say that in a 100g sample, we have 95 grams of ethanol and 5 grams of water

we will find the number of moles of ethanol and water in 100g solution of rectified spirit and use that to calculate the mole fraction

Moles of Ethanol:

Molar mass of ethanol = 46 grams / mol

Number of moles = Given mass / molar mass

Number of moles = 95 / 46

Moles of Ethanol = 2 moles (approx)

Moles of Water:

Molar mass of water = 18 grams per mol

Number of moles = Given mass / molar mass

Moles of water = 5 / 18

Moles of water  =  0.28 moles   (approx)

Mole Fractions:

Mole fraction of a specific compound is the number of moles of that compound divided by the total number of moles in the solution

Mole fraction of Ethanol:

Moles of ethanol / (moles of ethanol + moles of water)

2 / (2 + 0.28)

2 / (2.28) = 0.9 (approx)

Mole fraction of Water:

Moles of water / (Moles of ethanol + moles of water)

0.28 / (2 + 0.28)

0.28 / (2.28) = 0.1 (approx)

Answer:

Explanation:

Let the final volume be v .

from the formula

S₁ V₁ = S₂V₂

4 x 7 = .8 x V₂

V₂ = 35 mL

Final volume of the solution will be 35 mL .

Answer:

1.53 L

Explanation:

Step 1: Given data

Step 2: Calculate the moles (n) corresponding to 11.2 g of oxygen

The molar mass of oxygen is 32.00 g/mol.

11.2 g × (1 mol/32.00 g) = 0.350 mol

Step 3: Calculate the volume of oxygen

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 0.350 mol × (0.0821 atm.L/mol.K) × 415 K / 7.78 atm

V = 1.53 L

Fe(s) would have the greatest ductility.

Ductility is the capability of a fabric to be drawn or plastically deformed without fracture. it's far therefore a demonstration of how 'gentle' or malleable the fabric is. The ductility of steel varies relying on the sorts and levels of alloying factors gift.

Ductility is the property of metallic associated with the capability to be stretched into twine without breaking. Malleability is the assets of metallic associated with the ability to be hammered into thin sheets without breaking. The outside force or strain is tensile pressure.

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Answer:

Rubidium

Answer: Rubidium is the most reactive metal. Explanation: Metals are the elements that looses electrons and thus, their chemical reactivity will be the tendency to loose electrons.

Explanation:

Answer:

Phagocytosis

Explanation:

Answer:

[tex]x_{He}=0.975\\x_{O_2}=0.025[/tex]

Explanation:

Hello.

In this case, since we know the mass of both helium and oxygen, in order to obtain the mole fractions we first need the compute the moles by using their atomic masses, 4.00 g/mol and 32.00 g/mol respectively as shown below:

[tex]n_{He}=23.6gHe*\frac{1molHe}{4.00gHe}=5.90molHe\\ \\n_{O_2}=4.85gO_2*\frac{1molO_2}{32.00gO_2}=0.152molO_2\\[/tex]

Therefore, the mole fractions are:

[tex]x_{He}=\frac{n_{He}}{n_{He}+n_{O_2}}=\frac{5.90}{5.90+0.152} \\\\x_{He}=0.975\\\\x_{O_2}=\frac{n_{O_2}}{n_{He}+n_{O_2}} =\frac{0.152}{5.90+0.152} \\\\x_{O_2}=0.025[/tex]

Best regards!

Answer:

inorganic compounds that take the form of discrete molecules

Answer:

Compounds that take the form of discrete molecules.

Explanation:

Rather than forming ions, the atoms of a molecule share their electrons in such a way that a bond forms between pairs of atoms.

2x = 350- x

3x = 350

x = 350/x

therefore,

x = 116.67cm.

so, (350 - x)cm³ = (350- 166.67)cm³ = 233.33cm³

Hence distance covered by gas A is 116.67cm that by B is 233.33cm

Answer:

corbon 12

Explanation:

carbon atoms with the usual 6 nutrons have a mass number of 12

6 protons plus 7 neutrons equals 13 for carbon nuclei with 7 neutrons, these atoms make up the carbon-13 element.

A naturally occurring stable isotope of carbon with a nucleus comprising six protons and seven neutrons is carbon-13 (13C). It makes up roughly 1.1% of all the native carbon on Earth and is one of the environmental isotopes.

Since they are carbon atoms with extra neutrons, carbon 12, 13, and 14: You could also say that the difference in mass between these isotopes is molecular mass.

Therefore, molecular mass is determined by adding the number of protons and neutrons (if you have 1 mole of each isotope, carbon 14 would have the greatest mass).

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Answer:

Specific heat of aluminum  = 0.27 j/g.°C

Explanation:

Given data:

Mass of Al = 100 g

Initial temperature = 108.4°C

Final temperature = 68.2°C

Heat released = -1080 j

Specific heat of aluminum = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  68.2°C -  108.4°C

ΔT = - 40.2°C

Q = m.c. ΔT

-1080 j = 100 g ×c ×- 40.2°C

-1080 j = -4020 g.°C ×c

c = -1080 j/-4020 g.°C

c = 0.27 j/g.°C

Answer:

dangerous woman by ariana grande

Explanation:

Answer:

C6H10O6

Explanation:

Convert % and make them to grams. Convert grams to mols. Use mole ratios to find empirical formula. I will calculate it in a second. Hope this helps for now. This is my quick answer imma check my work one sec.

Answer:

true

Explanation:

hope it helps

Answer:

I'm pretty sure it's true

Explanation:

20 characters

Answer:

What mass of sodium hydroxide is needed to react completely with 10.0 g of iron( III) chloride? The conversion sequence is: A. B.

Explanation:

make me as brain liest

Answer:

Explanation:

The density of a substance can be found by using the formula

[tex]density = \frac{mass}{volume} \\ [/tex]

volume = final volume of water - initial volume of water

volume = 60 - 25 = 35 mL

From the question we have

[tex]density = \frac{140}{35} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

The molecular formula of the hydrocarbon is C10H10

Explanation:

Here, we are interested in finding the molecular formula for the hydrocarbon.

Since there are equal number of moles of carbon and hydrogen, then we have the molecular formula looking like;

CnHn

Kindly recall that the atomic mass of carbon is 12 a.m.u while that of hydrogen is 1 amu

so calculating the atomic mass of the compound, we have;

12(n) + 1(n) = 130.18

13n = 130.18

n = 130.18/13

n = 10.01

So the molecular formula will be C10H10

Answer:

C10H10

Explanation:

A hydrocarbon is a binary compound of carbon and hydrogen. Hence a hydrocarbon is a compound of the general formula (CH)n

Thus;

(12 + 1) n = 130.18

n= 130.18/13

n = 10

Hence the molecular formula of the compound is C10H10

Answer:

1 gram of N2 is equal to 0.035697202053303 mole. i dont know the rest but intried to help

. hopefully this will give you someone else a starting point. goodluck

Answer:

It's a representative metal

Explanation: Transitional metals are metals of various chemical elements and have valence electrons—i.e., electrons that can participate in the formation of chemical bonds.

Answer: Membrane Protiens

Explanation:

In facilitated diffusion, molecules diffuse across the plasma membrane with assistance from membrane proteins, such as channels and carriers. A concentration gradient exists for these molecules, so they have the potential to diffuse into (or out of) the cell by moving down it.

Answer:

Kindly check the explanation section.

Explanation:

Without mincing words let us dive right into the solution to the question above, taking each compound at a time.

NB: Kindly Check attachment for the Lewis Structure of each of the chemical compounds.

Therefore, the number of valence electrons, electronic shape, molecular shape and whether the molecules are polar(Polarity) is given below for each chemical compound.

(1). Compound: HCN

(a). number of valence electrons = 10.

(b). electronic shape =linear.

(c). molecular shape = linear.

(d). Polarity = Y.

(2). Compound: PH3

(a). number of valence electrons = 8.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Trigonal Pyramidal.

(d). Polarity = Y.

(3). Compound: CHCl3.

(a). number of valence electrons = 26.

(b). electronic shape = tetrahedral.

(c). molecular shape = tetrahedral.

(d). Polarity = Y.

(4). Compound: NH4^+

(a). number of valence electrons = 8

(b). electronic shape = tetrahedral

(c). molecular shape = tetrahedral

(d). Polarity = Y.

(5). Compound: H2CO

(a). number of valence electrons = 12.

(b). electronic shape = Trigonal planar.

(c). molecular shape = Trigonal planar

(d). Polarity = Y.

(6). Compound: SO4^2-

(a). number of valence electrons = 32.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Tetrahedral.

(d). Polarity = N.

(7). Compound: SeF2.

(a). number of valence electrons = 20.

(b). electronic shape = Tetrahedral.

(c). molecular shape = bent.

(d). Polarity = Y.

(8). Compound: CO2.

(a). number of valence electrons = 16.

(b). electronic shape = linear.

(c). molecular shape = linear.

(d). Polarity = N.

(9). Compound: O2

(a). number of valence electrons = 32.

(b). electronic shape = Trigonal planar.

(c). molecular shape = Linear.

(d). Polarity = N.

(10). Compound: ClO4-.

(a). number of valence electrons = 32.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Tetrahedral.

(d). Polarity = N.

(11). Compound: HBr.

(a). number of valence electrons = 8.

(b). electronic shape = Linear.

(c). molecular shape = Linear.

(d). Polarity = Y.

(12). Compound: PF5.

(a). number of valence electrons = 40.

(b). electronic shape = Trigonal Bipyramidal.

(c). molecular shape = Trigonal Bipyramidal.

(d). Polarity = N.

(13). Compound: BeH2.

(a). number of valence electrons = 4.

(b). electronic shape = Linear.

(c). molecular shape = Linear.

(d). Polarity = N.

(14). Compound: PO4^3-.

(a). number of valence electrons = 32.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Tetrahedral.

(d). Polarity = N.

(15). Compound: BH3.

(a). number of valence electrons = 6.

(b). electronic shape = Trigonal planar.

(c). molecular shape = Trigonal planar.

(d). Polarity = N

(16). Compound: Br3-.

(a). number of valence electrons = 32.

(b). electronic shape = Trigonal Bipyramidal.

(c). molecular shape = Linear.

(d). Polarity = N.

Answer:

The wallet is falling from a height of 1450 m

We are given that:

downward acceleration (a) = 10 m/s/s

downward displacement (s) = 1450 m

downward initial velocity (u) = 0 m/s (Since the wallet was dropped, it had no initial velocity)

horizontal initial velocity (v) = 13 m/s

Since there is no external force being applied on the wallet except the force of gravity, the wallet will keep moving at a constant velocity of 13 m/s

To calculate the horizontal distance travelled by the wallet, we need to know how long the wallet was airborne

To calculate how long the wallet was airborne, we have to find out that how long it took the wallet to hit the ground

Time taken for the wallet to hit the ground:

From the second equation of motion:

s = ut + 1/2 at² ------------(for vertical motion of the wallet)

1450 = (0)(t) + 1/2 (10)(t)²

1450 = 5t²

t² = 290

t = 17 (approx)

Time taken for the wallet to hit the ground = 17 seconds

Horizontal distance travelled by the wallet:

Since the wallet hits the ground after 17 seconds, it will move horizontally at a constant velocity of 13 m/s for 17 seconds

horizontal acceleration = 0 m/s/s

From the second equation of motion:

s = ut + 1/2at²

s = ut + 1/2 (0)t²

s = ut

here, u is the horizontal initial velocity of the wallet and the time taken by the wallet to hit the ground is 't'

s = (13)(17)

s = 221 m

Hence, the horizontal distance travelled by the wallet while falling from a height of 1450 m is 221 m

Answer:Two Main Types of Pure Substances

Elements and compounds are the two types of pure substances. Examples of common elements include carbon, nitrogen and hydrogen. They consist of one type of atom and cannot break down into something else. Every pure carbon substance, for example, has the same particles in it.

Explanation:I think

Answer:

3.76

Explanation:

You start by finding the mass of part d. Mass=  Density times Volume.In this case you would multiply 62.7 and 2.59 which is 162.393 rounded is 162.4. Then density= mass/ volume which is 162.4/ 43.2. density equals 3.76

Answer:

[tex]T=2.78x10^3 \°C[/tex]

Explanation:

Hello,

In this case, considering that the safe temperature may be computed via the ideal gas law as we now the pressure, mass and volume via the dimensions:

[tex]V=\pi r^2 h=\pi *(41.0cm)^2*49.2cm=2.60x10^5cm^3*\frac{1L}{1000cm^3} =260L[/tex]

The pressure in atm is:

[tex]P=3.70MPa*\frac{1x10^6Pa}{1MPa} \frac{1atm}{101325Pa} =36.5atm[/tex]

And the moles considering the mass and molar mass (66 g/mol) of dinitrogen difluoride (N₂F₂):

[tex]n_{N_2F_2}=2.50kg*\frac{1000g}{1kg}*\frac{1mol}{66g} =37.9mol[/tex]

In sich a way, by applying the ideal gas equation, which is not the best assumption but could work as an approximation due to the high temperature, the temperature, with three significant figures, will be:

[tex]T=\frac{PV}{nR}=\frac{36.5Pa*260L}{37.9mol*0.082\frac{atm*L}{mol*K} }\\ \\T=3053.6K-273.15\\\\T=2.78x10^3 \°C[/tex]

Best regards.

Explanation:

from

de Broglie wavelength= h/ mv

substitute the values

joules= mass(kg) × acceleration (m/s^2) × distance(m)= kgm^2/s^2.

sorry I don't have my calc with me

According to De Broglie's Hypothesis, matter has both particle and wave properties. The wavelength of De Broglie is given as λ = h p,

wavelength (in m) of a proton =  11.09 x[tex]10^ {-55[/tex] m.

λ= h m v.

= 6.63 x [tex]10^{-34}[/tex] x  1.673 x [tex]10^{-24[/tex]x 1000

= 11.09 x[tex]10^ {-55[/tex] m

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Answer:

Required heat = 2,259.36 N (approx)

Explanation:

Given:

Mass = 135 g

Specific heat (c) =4.184 J/gC

Change in temperature ΔT = 27-23 = 4 c

Find:

Required heat

Computation:

Q = mcΔT

Q = (135)(4.184)(4)

Required heat = 2,259.36 N (approx)

The heat transferred by the chemical reaction to the reservoir of a calorimeter containing 135 g of dilute aqueous solution is 2259.36 J

To solve this question, we'll begin by calculating the change in the temperature of the reservoir. This can be obtained as follow:

Initial temperature (T₁) = 23 °C

Final temperature (T₂) = 27 °C

ΔT = T₂ – T₁

ΔT = 27 – 23

Finally, we shall determine the heat transferred to the reservoir.

Change in temperature (ΔT) = 4 °C

Mass (M) = 135 g

Specific heat capacity (C) = 4.184 J/gºC

Q = MCΔT

Q = 135 × 4.184 × 4

Therefore, the heat transferred by the chemical reaction to the reservoir of a calorimeter containing 135 g of dilute aqueous solution is 2259.36 J

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