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2 years ago

What would be the mass, in grams, of 3.50 x 1025 molecules of chlorine? Show your work

Chemistry

1 answer:

Maurinko [17]2 years ago

4 0

Answer:

2059.645g Cl

Explanation:

A mole of any element contains 6.022e23 molecules, so we will divide the given amount of molecules by 6.022e23.

3.5e25 / 6.022e23 = 5.81e1

5.81e1 = 58.1 mol Cl

Now that we have found the amount of moles of Chlorine, we will multiply the number of moles by the molar mass of Chlorine, which is 35.45g/mol.

35.45(58.1) = 2059.645g Cl

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Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left

Vera_Pavlovna [14]

So C12 is the limiting reactant and P4 is the excess
Mass P4 consumed= 0.31 mol X 123.9 g/mol =38.41 g P4 consumed.
Hope i helped

6 0

3 years ago

I need help! It’s 10 point

pogonyaev

Answer:

$10^{5}$

Explanation:

Each value on the scale represents 10 times the previous value.

Subtract 1 from 6, so 10 to the power of 5 is the difference.

3 0

3 years ago

A student balanced the chemical equation Mg + O2 →MgO by writing Mg + O2 → MgO2. Was the equation balanced correctly? Explain yo

Amanda [17]

Explanation:

Charges on both magnesium and oxygen is 2. Though opposite in sign, they have equal charges so, both of them will be cancelled by each other.

As a result, formula of magnesium oxide is MgO and not $MgO_{2}$ .

The student write the equation as $Mg + O_2 \rightarrow MgO_2$ , it is not correct.

Therefore, given equation will be balanced as follows.

$2Mg + O_{2} \rightarrow 2MgO$

Since, number of atoms on both reactant and product side are equal. Hence, this equation is completely balanced.

7 0

3 years ago

What is the mass of 0.50 mole of copper?<br> 63.5 grams<br> O 32 grams<br> 3.2 grams<br> 2.18 grams

Amanda [17]

Answer:

32 g Cu

Explanation:

1 mol Cu -> 63.5 g

0.5 mol Cu ->x

x=(0.5 mol *63.5 g)/1 mol x= 32 g Cu

3 0

2 years ago

2h3po4 + 3mg(oh)2 → mg3(po4)2 + 6h2o phosphoric acid, h3po4, is neutralized by magnesium hydroxide, mg(oh)2, according to the eq

Rina8888 [55]

The balanced equation for the neutralisation reaction is as follows
2H₃PO₄ + 3Mg(OH)₂ --> Mg₃(PO₄)₂ + 6H₂O
stoichiometry of H₃PO₄ to H₂O is 2:6
number of H₃PO₄ moles reacted - 0.24 mol
if 2 mol of H₃PO₄ form 6 mol of H₂O
then 0.24 mol of H₃PO₄ forms - 6/2 x 0.24 = 0.72 mol of H₂O
therefore 0.72 mol of H₂O are formed

4 0

3 years ago