If the force acting upon an object is increased, what happens to the object’s acceleration?

How does temperature affect the vapor pressure of a liquid?

Mamont248 [21]

Answer:

As the temperature of a liquid increases, the kinetic energy of its molecules also increases. As the kinetic energy of the molecules increases, the number of molecules transitioning into a vapor also increases, thereby increasing the vapor pressure.

3 0

3 years ago

at 20°c, a 0.376 m aqueous solution of ammonium chloride has a density of 1.0045 g/ml. what is the mass % of ammonium chloride i

madam [21]

At 20°c, a 0.376 m aqueous solution of ammonium chloride has a density of 1.0045 g/ml. The mass percent of the ammonium chloride in the solution is 2.003%.

The number of moles of a substance can be estimated as:

$\mathbf{number\ of\ moles = \dfrac{mass}{molar \ mass}}$

Given that:

number of moles = 0.376 M
molar mass of ammonium chloride = 53.50 g/ml

Mass = number of moles × molar mass

Mass = 0.376 × 53.50

Mass = 20.116 grams

In one mL of ammonium chloride; the grams of ammonium chloride is:

$\mathbf{= \dfrac{20.116 \ grams}{1000 \ mL}}$

= 0.020116 g/mL

The mass of ammonium chloride in the solution can be computed as:

$\mathbf{=(\dfrac{mass \ of \ NH_4Cl}{density \ of \ the \ solution})\times 100}$

$\mathbf{=(\dfrac{0.020116 }{1.0045})\times 100}$

= 2.003%

Therefore, we can conclude that the mass percent of the ammonium chloride in the solution is 2.003%.

Learn more about ammonium chloride here:

brainly.com/question/23387600?referrer=searchResults

7 0

3 years ago

When comparing the two elements k and cs , the element with the more negative electron affinity is?

Nutka1998 [239]

The answer to this question would be: cs

Electron affinity is determined by the amount of energy released/spent for the atom to add an electron for neutral atoms in gas state.
Lower periodic table means a bigger radius atom. In this case, the distance between the electron to the nucleus is further too, so the affinity is expected to become lower.

In this question, the K and Cs are located in the first column of the periodic table, but Cs located lower. So, the Cs that located lower in periodic table should have a lower affinity.

4 0

3 years ago

A 25.00 mL sample containing an unknown amount of Al3+ and Pb2+ required 17.14 mL of 0.04907 M EDTA to reach the end point. A 50

dolphi86 [110]

Answer:

pAl³⁺ = 1,699

pPb²⁺ = 1,866

Explanation:

In this problem, the first titration with EDTA gives the moles of Al³⁺ and Pb²⁺, these moles are:

Al³⁺ + Pb²⁺ in 25,00mL = 0,04907M×0,01714L = 8,411x10⁻⁴ moles of EDTA≡ moles of Al³⁺ + Pb²⁺.

The molar concentration is 8,411x10⁻⁴ moles/0,02500L = 0,0336M Al³⁺+Pb²⁺.

In the second part each Al³⁺ reacts with F⁻ to form AlF₃. Thus, you will have in solution just Pb²⁺.

The moles added of EDTA are:

0,02500L×0,04907M = 1,227x10⁻³ moles of EDTA

The moles of EDTA in excess that react with Mn²⁺ are:

0,02064M × 0,0265L = 5,470x10⁻⁴ moles of Mn²⁺≡ moles of EDTA

That means that moles of EDTA that reacted with Pb²⁺ are:

1,227x10⁻³ moles - 5,470x10⁻⁴ moles = 6,800x10⁻⁴ moles of EDTA ≡ moles of Pb²⁺.

The molar concentration of Pb²⁺ is:

6,800x10⁻⁴mol/0,0500L = 0,0136 M Pb²⁺

Thus, molar concentration of Al³⁺ is:

0,0336M Al³⁺+Pb²⁺ - 0,0136 M Pb²⁺ = 0,0200M Al³⁺

pM is -log[M], thus pAl³⁺ and pPb²⁺ are:

pAl³⁺ = 1,699

pPb²⁺ = 1,866

I hope it helps!

3 0

3 years ago

Determine the oxidation number of Cl in each of the following species.Cl2O7AlCl4-Ba(ClO2)2CIF4+

DIA [1.3K]

These are four questons and four answers:

Answers:

1) 7⁺

2) 1⁻

3) 3⁺

4) 5⁺

Explanation:

Question 1) Cl₂O₇:

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

2×X + 7 (-2) = 0

2X - 14 = 0

2X = +14

X = +14 /2 = + 7

Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.

Question 2) AlCl₄⁻

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

1 (+3) + 4X = - 1

+3 + 4X = - 1

4X = - 1 - 3

4X = - 4

X = - 1

Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.

Question 3) Ba(ClO₂)₂

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

+2 + 2X + 4 (-2) = 0

2X +2 - 8 = 0

2X - 6 = 0

2X = +6

X = + 3

Conclusion: the oxidation number of Cl in Ba(ClO₂)₂ is 3⁺.

Question 4) CIF₄⁺

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

X + 4(-1) = +1

X - 4 = +1

X = +1 + 4

X = + 5

Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.

6 0

3 years ago