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2 years ago

5. The density of ethanol is 0.789 g/mL. Find the mass of a sample of ethanol that has a volume of 150.0 mL.

Chemistry

1 answer:

GenaCL600 [577]2 years ago

6 0

. 00526 g is mass of ethanol

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What is atom economy?

bonufazy [111]

Answer:

The conversion efficiency of a chemical process.

Explanation:

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5 0

3 years ago

PLEASE HELPPPP!!!!!

Likurg_2 [28]

Answer : The mass of nitric acid is, 214.234 grams.

Solution : Given,

Moles of nitric acid = 3.4 moles

Molar mass of nitric acid = 63.01 g/mole

Formula used :

$\text{Mass of }HNO_3=\text{Moles of }HNO_3\times \text{Molar mass of }HNO_3$

Now put all the given values in this formula, we get the mass of nitric acid.

$\text{Mass of }HNO_3=(3.4moles)\times (63..01g/mole)=214.234g$

Therefore, the mass of nitric acid is, 214.234 grams.

6 0

3 years ago

Please Help !! Brainliest !!

Rufina [12.5K]

Answer:

Data supports significantly because we can use the testing(depending sample) before and after we use the same object to test the hypothesis.

5 0

3 years ago

When a piece of solid phosphorus is added to a flask containing chlorine gas, heat is evolved?

Sav [38]

White phosphorus burns spontaneously in chlorine to produce a mixture of two chlorides, phosphorus (ii) chloride and phosphorus(v) chloride and heat is evolved. Phosphorus (iii) chloride is a colorless fuming liquid:
P4 + 6Cl2 = 4PCl3
Phosphorus (v) chloride is an off-white solid
P4 +10Cl2 = 4PCl5

7 0

2 years ago

c. The reaction Br2 (l) --> Br2 (g) has ΔH = 30.91 kJ/mol and ΔS = 93.3 J/mol·K. Use this information to show (within close a

egoroff_w [7]

Answer:

The answer to your question is given below.

Explanation:

From the question given above, the following data were obtained:

Br₂ (l) —> Br₂(g)

Enthalpy change (ΔH) = 30.91 KJ/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

Next, we shall convert 30.91 KJ/mol to J/mol. This can be obtained as follow:

1 KJ/mol = 1000 J/mol

Therefore,

30.91 KJ/mol = 30.91 × 1000

30.91 KJ/mol = 30910 J/mol

Thus, 30.91 KJ/mol is equivalent to 30910 J/mol.

Finally, we shall determine the boiling temperature of bromine. This can be obtained as follow:

Enthalpy change (ΔH) = 30910 J/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

ΔS = ΔH / T

93.3 = 30910 / T

Cross multiply

93.3 × T = 30910

Divide both side by 93.3

T = 30910 / 93.3

T = 331.29 K

Thus, the boiling temperature of bromine is 331.29 K

6 0

2 years ago