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JulijaS [17]
3 years ago
6

Does the +2 and -1 make a difference when solving for the Hrxn? What difference does it make?

Chemistry
1 answer:
Kruka [31]3 years ago
4 0

Answer:

It would be a square plus b square and i am writing just so i can get pionts

Explanation:

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How many moles of atoms are in 6.00 g of 13C?
Lyrx [107]

Answer: 0.462 moles

Explanation: 13C indicates an isotope of carbon and its mass number is 13. It means the mass of 1 mol of 13C is 13 gram.

The question asks to calculate the number of atoms present in 6.00 grams of 13C.

To calculate the number of moles we divide the given grams by the mass of 1 mol of the element. The set could be shown easily using dimensional analysis as:

6.00gram(\frac{1mol}{13gram})

= 0.462 moles

So, there will be 0.462 moles of atoms in 6.00 grams of 13C.

4 0
3 years ago
Read 2 more answers
Good Morning I have a question I need help and can not find the answer o it maybe someone help me? The question is _______ Most
Semenov [28]

Answer:

hydro, water

Explanation:

3 0
3 years ago
You have 125.0 mL of a solution of H3PO4, bu you don't know its concentration. if you titrate the solution with a 4.56-M solutio
irakobra [83]

Answer:

4.90 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 125.0 mL

M₂ = 4.56 M

V₂ = 134.1 mL

Using the above formula , the molarity of acid , can be calculated as ,

M₁V₁ = M₂V₂  

Substituting the respective values ,  

M₁ *  125.0 mL = 4.56 M *  134.1 mL

M₁ = 4.90 M

6 0
4 years ago
A saturated hydrocarbon is an organic compound which contains hydrogen and carbon joined _____.
Leya [2.2K]
B) with single bonds..... Hope it helps, Have a nice day :)
8 0
3 years ago
Read 2 more answers
15. What volume of CCI, (d = 1.6 g/cc) contain
anastassius [24]

Answer:

\boxed{\text{(3) 9.6 L}}

Explanation:

1. Moles of CCl₄

n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}

4. Volume of CCl₄

V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}

4 0
3 years ago
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