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EleoNora [17]
3 years ago
14

If you know the number of protons what else do you know the number of?

Chemistry
2 answers:
pav-90 [236]3 years ago
5 0
You know the atomic number
Vikki [24]3 years ago
5 0
<span>If you know the number of protons you also know the number of electrons because an element has the same number of electrons as it does protons.</span>
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Which best describes the relationship between Haiti’s political instability and the natural disasters it has faced?
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A scientist wants to perform a controlled experiment to test the effect of rust on chemically treated iron. which action would b
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The answer is to test a piece of untreated and treated under same conditions!
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PLEASE HELP!!!!!!!
dsp73

Answer: 1. 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

2. 3 moles of CuCl_2 : 2 moles of Al

3. 0.33 moles of CuCl_2 : 0.92 moles of Al

4. CuCl_2 is the limiting reagent and Al is the excess reagent.

5. Theoretical yield of AlCl_3 is 29.3 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles

\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles

The balanced chemical equation is:

2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu  

According to stoichiometry :

3 moles of CuCl_2 require = 2 moles of Al

Thus 0.33 moles of CuCl_2 will require=\frac{2}{3}\times 0.33=0.22moles  of Al

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Al is the excess reagent.

As 3 moles of CuCl_2 give = 2 moles of AlCl_3

Thus 0.33 moles of CuCl_2 give =\frac{2}{3}\times 0.33=0.22moles  of AlCl_3

Theoretical yield of AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3

Thus 29.3 g of aluminium chloride is formed.

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2 years ago
PPLEASE HELP I WILL GIVE BRAINLIEST AND LOT OF POINTS
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6 0
3 years ago
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A flask with a volume of 3.16 l contains 9.33 grams of an unknown gas at 32.0°c and 1.00 atm. What is the molar mass of the gas?
Inessa05 [86]

Answer:

73.88 g/mol

Explanation:

For this question we have to keep in mind that the unknown substance is a <u>gas</u>, therefore we can use the <u>ideal gas law</u>:

PV=nRT

In this case we will have:

P= 1 atm

V= 3.16 L

T = 32 ªC = 305.15 ºK

R= 0.082 \frac{atm*L}{mol*K}

n= ?

So, we can <u>solve for "n"</u> (moles):

1~atm*3.16~L~=~n*0.082~\frac{atm*L}{mol*K}*305.15~K

n=\frac{1~atm*3.16~L~}{0.082~\frac{atm*L}{mol*K}*305.15~K}

n=0.126~mol

Now, we have to remember that the <u>molar mass value has "g/mol"</u> units. We already have the grams (9.33 g), so we have to <u>divide</u> by the moles:

molar~mass=\frac{9.33~grams}{0.126~mol}

molar~mass=73.88\frac{grams}{mol}

7 0
3 years ago
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