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Maurinko [17]
3 years ago
9

Join google classroom with the code given in picture below

Chemistry
2 answers:
lana [24]3 years ago
6 0

Answer: I guess?

Explanation:

gladu [14]3 years ago
5 0

Answer:

what is this for

Explanation:

what is this for

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The anion of which element has a –1 charge?
vovangra [49]
An anion is a negatively charged ion. An element can become an anion if it takes an extra electron such that it has one more electron than protons.
Atoms would preferentially taken an extra electron so that it can have a full octet and be more stable.

\sf Cl^-, Br^-, F^- are some of the few elements that would have a -1 charge.
6 0
3 years ago
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

Therefore the percent yield is:

Y=\frac{112g}{285.13g}*100\%\\\\Y=39.3\%

Best regards!

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