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AnnyKZ [126]
3 years ago
13

I need help with #17 from my Chemistry homework...

Chemistry
1 answer:
ohaa [14]3 years ago
7 0
Hand it in to your teacher or professor like that he or she will give you an A+
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Is respiration like burning?Explain your answer.​
SIZIF [17.4K]

Answer:

Yes

Explanation:

The process of respiration and burning are similar in the following ways: Both respiration and burning require oxygen. Energy is released during both respiration and burning. The products produced (carbon dioxide and water) are the same for both respiration and burning.

3 0
2 years ago
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How many moles are in 8.25×10²⁵ molecules of water(H20)?​
enyata [817]

Answer:

82500000000000000000000000

Explanation:

This is the only answer I can come up with.

6 0
2 years ago
A solution is prepared by adding 100 ml of 0.2 m hydrochloric acid to 100 ml of 0.4 m sodium formate. is this a buffer solution,
Firdavs [7]
Becoing itsodim fromate t will have  pb of 67

5 0
3 years ago
Flammability is a material’s ability to burn in the presence of
frosja888 [35]

Answer:

Flammability is a material’s ability to burn in the presence of <u><em>oxygen.</em></u>

Explanation:

Flammability can be described as the ability of a substance to get ignited. Flammability will lead to fire or combustion. Some substances are highly flammable like Benzene. Other tend to be just flammable. And there are also compounds which will nor be flammable at all as they won't react with oxygen. Examples of these substances include helium, steel or glass.

The flammability of a substance shall be considered a very important aspect when storing or transporting a substance.

5 0
3 years ago
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.

5 0
2 years ago
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