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3 years ago

If 0.45 mols of MnO2 can react with 0.28 moles of HCl, how many moles of Cl2 could be produced

1 answer:

3 0

MnO₂ + 4HCl = MnCl₂ + 2H₂O + Cl₂

n(MnO₂)=0.45 mols
n(HCl)=0.28 mols

0.45 : 0.28 = 1 : 0.622
the reaction equation 1:4
the hydrogen chloride in the lack

n(Cl₂)=n(HCl)/4
n(Cl₂)=0.28/4=0.07 mols
formed 0.07 mols of chlorine

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In which mutation does one base take the place of another?

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Answer: A substitution is a mutation that exchanges one base for another (i.e., a change in a single "chemical letter" such as switching an A to a G). Such a substitution could: change a codon to one that encodes a different amino acid and cause a small change in the protein produced.

Explanation: I know this is correct just trust me and please mark me as brainiest.

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3 years ago

Perform the calculation and record the answer with the correct number of significant figures.

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The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

Answer:

For A: The answer becomes 0.1

For B: The answer becomes 0.4884

Explanation:

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

Rule applied for addition and subtraction:

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

Rule applied for multiplication and division:

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

For A: (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

$\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}$

Here, the least precise number after decimal was 1.

$\Rightarrow \frac{0.4}{3.19}=0.125$

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

For B: (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

$\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}$

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

$\Rightarrow \frac{43.72}{89.516}=0.48840$

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

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3 years ago

One of the commercial uses of sulfuric acid is the production of calcium sulfate and phosphoric acid. If 26.8 g of Ca₃(PO₄)₂ rea

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Answer:

The percent yield reaction is 64.3%

Explanation:

This is the ballanced reaction

Ca₃(PO₄)₂ (s) + 3H₂SO₄ (aq) → 2H₃PO₄ (aq) + 3CaSO₄ (aq)

Let's determine the moles of our reactants:

Mass / Molar mass = Mol

26.8 g / 310.18 g/m = 0.0864 moles of phosphate.

54.3 g / 98.06 g/m = 0.554 moles of sulfuric

1 mol of phosphate reacts with 3 mol of sulfuric so

0.0864 mol of PO₄⁻³ will react with (0.0864 .3)/1 = 0.259 moles

I have 0.554 of sulfuric, so this is the reactant in excess.

The limiting reagent is the Phosphate.

1 mol of PO₄⁻³ produces 2 mol of phosphoric

0.0864 of PO₄⁻³ will produce the double amount (0.0864 .2) = 0.173 moles

Mol . molar mass = Mass

0.173 m . 97.98g/m = 16.95 g (This is the theoretical yield)

Percent yield = (Produced / Theoretical) .100

(10.9 g / 16.95 g) . 100 = 64.3 %

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