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elena-s [515]
3 years ago
14

#1: A material has a density of 8.9 g/cm^3. You have six cubic centimeters of the substance. What is the material’s mass in gram

s?
A. 3.0 g

b. 5.9 g

C. 27 g

D. 53 g
**Idk i think it is either C. 27 g or D. 53 g :/
Chemistry
2 answers:
Airida [17]3 years ago
5 0

Answer:

53g

Explanation:

done the topic

sveta [45]3 years ago
3 0
<span>The density of a material = mass/ volume From the question, volume = 6 cm^3. Since the density = 8.9 g/cm^3 We have that 8.9 = mass/ 6 So mass = 8.9 * 6 = 53.4 So it follows that our mass = 53.4g. Hence option D which is 53g.</span>
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Answer:

2Cu + O2 ----------------> 2CuO

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5 0
3 years ago
Please answer, with explanation. Thanks!​
nadya68 [22]

Answer:

Explanation:

a = 40.1 g of Ca

Number of moles = mass / molar mass

Number of moles = 40.1 g/ 40.1 g/mol

Number of moles = 1 mol

b = 11.5 g Na

Number of moles = mass / molar mass

Number of moles = 11.5 g/ 23 g/mol

Number of moles = 0.5 mol

c = 5.87 g Ni

Number of moles = mass / molar mass

Number of moles = 5.87 g/ 58.7 g/mol

Number of moles = 0.1 mol

d = 150 g of S

Number of moles = mass / molar mass

Number of moles = 150 g/ 32 g/mol

Number of moles = 4.7 mol

e = 2.65 g Fe

Number of moles = mass / molar mass

Number of moles = 2.65 g/ 55.85 g/mol

Number of moles = 0.05 mol

f = 0.00750 g Ag

Number of moles = mass / molar mass

Number of moles = 0.00750 g/ 107.9 g/mol

Number of moles = 6.95 × 10⁻⁵ mol

g = 2.25 × 10²⁵ atoms Zn

1 mole = 6.022 × 10²³ atoms

1 mol / 6.022 × 10²³ atoms × 2.25 × 10²⁵ atoms

0.17  × 2.25 × 10²⁵ moles

38.25 moles

h = 50 atoms of Ba

1 mole = 6.022 × 10²³ atoms

1 mol / 6.022 × 10²³ atoms ×50 atoms

0.17 × 10²³ × 50 moles

8.5 × 10²³ moles

6 0
3 years ago
An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx
Anna [14]

Answer:

M=235.42g/mol

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

2n_{acid}=n_{base}

It means that the moles of acid can be computed given the volume and concentration of NaOH:

n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol

It means that the approximate molar mass of the acid is:

M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol

Best regards!

5 0
3 years ago
What is the expected value for the heat of sublimation of acetic acid if its heat of fusion is 10.8 kJ/mol and its heat of vapor
Dennis_Churaev [7]

Answer:

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

Explanation:

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Heat of vaporization of acetic acid = H^o_{vap}=24.3 kJ/mol

CH_3COOH(s)\rightarrow CH_3COOH(l)..[2]

Heat of fusion of acetic acid = H^o_{fus}=10.8 kJ/mol

Heat of sublimation of acetic acid = H^o_{sub}=?

CH_3COOH(s)\rightarrow CH_3COOH(g)..[3]

[1] + [2] = [3] (Hess's law)

H^o_{sub}=H^o_{vap}+H^o_{fus}

=24.3 kJ/mol+10.8 kJ/mol=35.1 kJ/mol

35.1 kJ/mol is the expected value for the heat of sublimation of acetic acid.

5 0
3 years ago
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