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3 years ago

What is the answer, with the correct number of significant figures, for this problem? 4.392g + 102.40g + 2.51g =_______.

Chemistry

1 answer:

MrRa [10]3 years ago

7 0

Answer : The correct answer will be, $109.30g$

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

The rule apply for the addition and subtraction is :

The least precise number present after the decimal point determines the number of significant figures in the answer.

The given expression is:

$4.392g+102.40g+2.51g$

$\Rightarrow 109.302g$

In the given expression, 4.392 has 4 significant figures, 102.40 has 5 significant figures and 2.51 has 2 significant figures. From this we conclude that least precise number present after the decimal point is 2.

Thus, the answer will be $109.30g$

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Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8). 2. What is the pH of a solution containing a hydrogen ion concentrat

Pepsi [2]

Answer:

$\large \boxed{1. \text{ 0.17 g/L; 2. 3.52; 3. Cl; 4. (a) +3; (b) +4; (c) +6}}$

Explanation:

1. Solubility of CaF_2

(a) Molar solubility

CaF₂ ⇌ Ca²⁺ + 2F⁻

$K_{\text{sp }} = \text{[Ca$^{2+}$]}\text{[F$^{-}$]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}$

(b) Mass solubility

$\text{Solubility} = 2.2 \times 10^{-3} \text{ mol/L} \times \dfrac{\text{78.07 g}}{\text{1 L }} = \text{0.17 g/L}\\\\\text{The solubility of CaF$_{2}$ is $\large \boxed{\textbf{0.17 g/L}}$}$

2. pH

pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52

3. Oxidizing and reducing agents

Zn + Cl₂ ⟶ ZnCl₂

$\rm \stackrel{\hbox{0}}{\hbox{Zn}} + \stackrel{\hbox{0}}{\hbox{ Cl}_{2} }\longrightarrow \stackrel{\hbox{+2}}{\hbox{Zn}}\stackrel{\hbox{-1}}{\hbox{Cl}_{2}}$

The oxidation number of Cl has decreased from 0 to -1.

Cl has been reduced, so Cl is the oxidizing agent.

4. Oxidation numbers

(a) Al₂O₃

$\stackrel{\hbox{$\mathbf{+3}$}}{\hbox{Al}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{3}}$

1O = -2; 3O = -6; 2Al = +6; 1Al = +3

(b) XeF₄

$\stackrel{\hbox{$\mathbf{+4}$}}{\hbox{Xe}}\stackrel{\hbox{-1}}{\hbox{F}_{4}}$

1F = -1; 4F = -4; 1 Xe = +4

(c) K₂Cr₂O₇

$\stackrel{\hbox{${+1}$}}{\hbox{K}_{2}}\stackrel{\hbox{$\mathbf{+6}$}}{\hbox{Cr}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{7}}$

1K = +1; 2K = +2; 1O = -2; 7O = -14

+2 - 14 = -12

2Cr = + 12; 1 Cr = +6

8 0

4 years ago

7. Fill in the chart with information on the following atoms.

erastovalidia [21]

Answer:

1016

1916

1020

Explanation:

3 0

3 years ago

A certain compound has the percent composition (by mass) 85.63% C and 14.37% H. The molar mass of the compound is 42.0 g/mol. Ca

AlekseyPX

Answer:

The molecular formula is C3H6

Explanation:

Step 1: Data given

Suppose the compound has a mass of 100 grams

The compound contains:

85.63 % C = 85.63 grams C

14.37 % H = 14.37 grams H

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles

Moles = grams / molar mass

Moles C = 85.63 grams / 12.01 g/mol

Moles C = 7.130 moles

Moles H = 14.37 grams / 1.01 g/mol

Moles H = 14.2 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 7.130 moles / 7.130 moles = 1

H = 14.2 moles / 7.130 moles = 2

The empirical formula is CH2

The molar mass of CH2 = 14 g/mol

Step 4: Calculate molecular formula

We have to multiply the empirical formula by n

n = 42 / 14 = 3

n*(CH2) = C3H6

The molecular formula is C3H6

8 0

3 years ago

95. Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulfur double bond in C

prohojiy [21]

Answer:

+523 kJ.

Explanation:

The following data will be used to calculate the average C-S bond energy in CS2(l).

S(s) ---> S(g)

ΔH = 223 kJ/mol

C(s) ---> C(g)

ΔH = 715 kJ/mol

Enthalpy of formation of CS2(l)

ΔH = 88 kJ/mol

CS2(l) ---> CS2(g)

ΔH = 27 kJ/mol

CS2(g) --> C(g) + 2S(g)

So we must construct it stepwise.

1: C(s) ---> C(g) ΔH = 715 kJ

2: 2S(s) ---> 2S(g) ΔH = 446 kJ

adding 1 + 2 = 3

ΔH = 715 + 446

= 1161 kJ

3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ

4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ

adding (reversed 3) from 4 = 5

ΔH = -1161 + 88

= -1073 kJ

5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ

6: CS2(l) ---> CS2(g) ΔH = 27 kJ

adding 5 + 6 = 7

ΔH = -1073 + 27

= -1046 kJ

7. C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ

Reverse and divide by 2 for C-S bond enthalpy

= -(-1046)/2

= +523 kJ.

8 0

3 years ago

A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s

Lilit [14]

Answer:

pH = 1.32

Explanation:

H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M

54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

3 0

3 years ago