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disa [49]
3 years ago
6

What is the molarity of a solution containing 6.0 moles of solute in 549 mL of solution?

Chemistry
2 answers:
Alexus [3.1K]3 years ago
7 0

b because of the solution

GrogVix [38]3 years ago
6 0

The answer is B. Hope that helps!

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Making methanol the element hydrogen is not abundant in nature, but it is a useful reagent in, for example, the potential synthe
stealth61 [152]
ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So 
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K
5 0
2 years ago
What is the boiling point, melting point, and triple point of bromine?
Lapatulllka [165]

Answer:

A. Boiling point = 59 °C, Melting point = -7.2°C, triple point = -7.3°C

Explanation:

3 0
3 years ago
A sample of a substance has a mass of 4.2 grams and a volume of 6 milliliters. The density of this substance is
Tanzania [10]

Hello!

Mass =4.2 g

Volume =6 mL

Therefore:

Density = mass / volume

Density = 4.2 / 6

Density = 0.7 g/mL


Hope that helps!

4 0
3 years ago
Read 2 more answers
57. A recipe for s'mores calls for 2 graham crackers, 2 chocolate pieces and 1 marshmallow. If I
geniusboy [140]

Answer:Noble gases:

 are highly reactive.

 react only with other gases.

 do not appear in the periodic table.

 are not very reactive with other elements.

Explanation:Noble gases:

 are highly reactive.

 react only with other gases.

 do not appear in the periodic table.

 are not very reactive with other elements.

5 0
2 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
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