Please help me #6!!!!!!!!

Chemistry

1 answer:

Vesnalui [34]3 years ago

3 0

I believe that the answer is 12 because there is already 3 O molecules and since its in parentheses with 3 outside it that means that there are 3 of those CO molecules meaning that for every 1 CO there will be 3 O’s so 3, four times Is 12

You might be interested in

How many oxygen atoms are present in 3 grams of glucose

Kaylis [27]

Answer:

Because CLEARLY, each mole of glucose, C6H12O6 contains 6⋅mol oxygen atoms.

5 0

3 years ago

How can we fix the affects the affects of acid rain

dybincka [34]

Adaptation actually and also following control measures on how to avoid it from happening

6 0

2 years ago

Bromine has the atomic number 35. Bromine atoms often have 36 electrons, making them have this overall charge:

Afina-wow [57]

Answer: A) -1

Explanation:

7 0

3 years ago

Read 2 more answers

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)a. Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of

Ira Lisetskai [31]

Answer:

a. 167 mL.

b. 39.3 %.

Explanation:

Hello!

In this case, for the undergoing chemical reaction, since 45.0 g of aluminum react, based on the 2:3 mole ratio with sulfuric acid, we can compute the required moles as shown below:

$n_{H_2SO_4}=45.0gAl*\frac{1molAl}{27.0gAl} *\frac{3molH_2SO_4}{2molAl} =2.50molH_2SO_4$

Next, since the molarity of a solution is computed based on the moles and volume (M=n/V), we can compute the required volume of sulfuric acid as shown below:

$V=\frac{n}{M}=\frac{2.50mol}{15.0mol/L}=0.167L$

That in mL is 167 mL.

Moreover, for the percent yield, we compute the grams of aluminum sulfate that are produced based on the required 2.50 moles of sulfuric acid:

$m_{Al_2(SO_4)_3}=2.50molH_2SO_4*\frac{1molAl_2(SO_4)_3}{3molH_2SO_4}*\frac{342.15gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=285.13gAl_2(SO_4)_3$