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Jet001 [13]
2 years ago
15

Below is a word equation. Please follow directions for each part below.

Chemistry
1 answer:
alukav5142 [94]2 years ago
8 0

Answer:

CH3OH + 02 ----> C02 + H20

balanced equation -

CH3OH + 3/202 ----> C02 + 2H20

Use exactly the same process as the one used on another question of yours I answered.

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The empirical formula for a compound that is 1.2% h, 42.0%cl, and 56.8%o is
Charra [1.4K]
Hope you understand how to work out those types of questions now xD ;)

7 0
3 years ago
A solution of hydrocloric acid has a molarity of 2.25 M HCl. If a reaction requires 5.80 g of HCI, what
Natali [406]

Answer:

0.071L

Explanation:

From the question given, we obtained the following data:

Molarity of HCl = 2.25 M

Mass of HCl = 5.80g

Molar Mass of HCl = 36.45g/mol

Number of mole of HCl =?

Number of mole = Mass /Molar Mass

Number of mole of HCl = 5.8/36.45 = 0.159mole

Now, we can obtain the volume required as follows:

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.159mole/ 2.25

Volume = 0.071L

4 0
4 years ago
How can constraints be used to help define the problem?
vitfil [10]

Answer:

Constraints are restrictions that need to be placed upon variables 

Explanation:

Constraints are restrictions (limitations, boundaries) that need to be placed upon variables used in equations that model real-world situations. It is possible that certain solutions which make an equation true mathematically, may not make any sense in the context of a real-world word problem.

6 0
3 years ago
If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?
kodGreya [7K]

Answer:

[Cu^{2+}]=0.041 M

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}

[Cu^{2+}]=0.041 M

Best regards!

6 0
3 years ago
To use the right hand rule, you should point your thumb in the same
Flauer [41]
The Correct Answer is Eletric Current I think
5 0
4 years ago
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