Charlie measures the mass of a metal three times and recorded 8.83 g,

Consider the following reaction. Zn(s) 2agno3(aq) --> 2ag(s) zn(no3)2(aq)when 16. 2 g of silver was produced, _____ mole(s) o

Crazy boy [7]

The required amount of silver nitrate to produce 16.2g of silver is 25.48 grams.

<h3>What is the relation between mass & moles?</h3>

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

W = given mass
M = molar mass

Moles of silver = 16.2g / 107.8g/mol = 0.15mol

From the stoichiometry of the given reaction it is clear that, same moles of silver nitrate is required to produce same moles of silver. So 0.15 moles of silver nitrate is required.

Mass of silver nitrate = (0.15mol)(169.87g/mol) = 25.48g

Hence required mass of silver nitrate is 25.48g.

To know more about mass & moles, visit the below link:

brainly.com/question/19784089

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5 0

2 years ago

Calculate how many grams of oxygen form when each quantity of reactant completely reacts.. . 2HgO----->2Hg+O2

goblinko [34]

<span>6.96g HgO
hope it helps.
</span>

3 0

3 years ago

15-9+6+3 =? What's the answer of this?

julia-pushkina [17]

15-9+6+3=

ANSWER

15

6 0

3 years ago

Help me plsssssssss Brainliest to correct answer

Elina [12.6K]

It may be unreliable because it might mess something up from the speed.

8 0

3 years ago

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

$n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC$

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

$n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH$

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

$m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO$

4. Compute the moles of oxygen by using its molar mass:

$n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO$

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

$C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1$

6. Search for the closest whole number (in this case multiply by 2):

$C_3H_6O_2$

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

$M=12*3+1*6+16*2=74g/mol$

Which is about three times in the molecular formula, for that reason, the actual formula is:

$C_9H_{18}O_6$

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0

3 years ago