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1 year ago

2HgO(s) 2Hg(l) + O2(g)

Chemistry

1 answer:

WITCHER [35]1 year ago

5 0

The theoretical yield of the Hg is 4.02 g.

<h3>What s the theoretical yield?</h3>

The theoretical yield is the yield that is obtained from the stoichiometry of the reaction.

The reaction equation is; 2HgO(s) ------> 2Hg(l) + O2(g)

Number of moles of HgO = 4.37 g/217 g/mol = 0.02 moles

Number of moles of Hg = 3.21 g/201 g/mol = 0.012 moles

Now;

Theoretical yield of Hg = 0.02 moles * 201 g/mol = 4.02 g

Learn more about theoretical yield:brainly.com/question/14966377

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You need to make an aqueous solution of 0.207 M calcium acetate for an experiment in lab, using a 300 mL volumetric flask. How m

jeka94

Answer:

We have to add 9.82 grams of calcium acetate

Explanation:

Step 1: Data given

Molarity of the calcium acetate solution = 0.207 M

Volume = 300 mL = 0.300 L

Molar mass calcium acetate = 158.17 g/mol

Step 2: Calculate moles calcium acetate

Moles calcium acetate = molarity * volume

Moles calcium acetate = 0.207 M * 0.300 L

Moles calcium acetate = 0.0621 moles

Step 3: Calculate mass calcium acetate

Mass calcium acetate = moles * molar mass

Mass calcium acetate = 0.0621 moles * 158.17 g/mol

Mass calcium acetate = 9.82 grams

We have to add 9.82 grams of calcium acetate

5 0

3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

vladimir1956 [14]

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ $\frac{5.60}{2}gallons = 2.80 gallons$

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = $(M__1}) = Density*Volume$

$= 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}$

$= 11659.06grams$

On the other hand; the mass of water = $(M__2})= Density*Volume$

$= 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}$

$= 10577.95 grams$

Molarity = $\frac{massof solute*1000}{molarmassof solute*massofwater}$

= $\frac{11659.06*1000}{62.07*10577.95}$

= 17.757 m

≅ 17.76 m

∴ the boiling point of the solution is calculated using the boiling‑point elevation constant for water and the Molarity.

$\Delta T_{boiling} = k_{boiling}M$

where,

$k_{boiling}$ = 0.512 °C/m

$\Delta T_{boiling}$ = 100°C + 17.56 × 0.512

= 109.09 °C

6 0

3 years ago

4 NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Mariulka [41]

Answer:

0.9 moles of water

Explanation:

Use mole ratios:

5 : 6

divide by 5 on both sides

1 : 1.2

multiply by 0.75 on both sides

0.75 : 0.9

So the result is 0.9 moles of water

(Please correct me if I'm wrong)

5 0

2 years ago

The empirical formula of a compound is CH2O and its molecular weight is 90.09 g/mol. Determine the molecular formula.

Pachacha [2.7K]

Answer : The molecular formula of the compound will be, $C_{3}H_{6}O_3$

Explanation :

Empirical formula : It is the simplest form of the chemical formula which depicts the whole number of atoms of each element present in the compound.

Molecular formula : it is the chemical formula which depicts the actual number of atoms of each element present in the compound.

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

As we are given that the empirical formula of a compound is $CH_2O$ and the molar mass of compound is, 90.09 gram/mol.

The empirical mass of $CH_2O$ = 1(12) + 2(1) + 1(16) = 30 g/eq

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

$n=\frac{90.09}{30}=3$

Molecular formula = $(CH_2O)_n=(CH_2O)_3=C_{3}H_{6}O_3$

Thus, the molecular formula of the compound will be, $C_{3}H_{6}O_3$

7 0

3 years ago

If it takes 5 hours to produce 8.0 kg of alcohol, how any days will it take to consume 1000 kg of glucose?

Ipatiy [6.2K]

It takes 21.3 days

<h3>Further explanation</h3>

Given

5 hr = 8 kg Alcohol

Required

Days to consume 1000 kg of glucose

Solution

Alcoholic fermentation⇒ glucose produces ethanol and carbon dioxide,

C₆H₁₂O₆ → 2 C₂H₅OH + 2CO₂

mol ethanol :

$\tt \dfrac{5000~g}{46~g/mol}=108.7~moles$

moles of glucose to produce 108.7 moles ethanol :

$\tt \dfrac{1}{2}\times 108.7=54.35$

54.35 moles = 5 hours

moles of 1000 kg of glucose :

$\tt \dfrac{10^6~g}{180~g/mol}=5555.5~moles$

So for 5555.5 moles, it takes :

$\tt \dfrac{5555.5}{54.35}\times 5~hours=511.085~h=21.3~days$

7 0

3 years ago