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2 years ago

How many moles are in 125g of water?

1 answer:

7 0

To calculate number of moles, all you do is divide the given mass by the molecular molar mass:

i.e. 125g / 18g = 6.94444g 

Therefore, your answer is (a) 6.94 g

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What tools do scientist use

Nikolay [14]

A beaker and a microscope

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2 years ago

Read 2 more answers

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

3 years ago

Could anyone give me the answer to this plz 20 points

svetoff [14.1K]

13 - Periodic table
14 - Dimitri mandeleev
15 - groups

Mark me brainiest pls it right answer

6 0

3 years ago

What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the overall r

Sergeu [11.5K]

Answer:
5 moles of electrons

Explanation:

The balance equation is as follow,

 5 Ag⁺ + Mn⁺² + 4 H₂O → 5 Ag + MnO₄⁻ + 8 H⁺

Reduction of Ag:

Ag⁺ + 1 e⁻ → Ag
Or,
5 Ag⁺ + 5 e⁻ → 5 Ag

Oxidation of Mn:

Mn⁺² → MnO₄⁻ + 5 e⁻

Result:
Hence 5 moles of Ag⁺ accepts 5 electrons from 1 mole of Mn⁺².

4 0

3 years ago

Which compound has the LEAST percent composition by mass of sulfur?

Eddi Din [679]

Answer: srs with 32.38 grams

Explanation:

5 0

2 years ago