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ycow [4]
4 years ago
7

Which determines the state of matter for any material?

Chemistry
1 answer:
Paul [167]4 years ago
8 0

Answer: the answer is A and C

Explanation: because they are really kind of both the same

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The boiling process is considered to be____
Dmitry_Shevchenko [17]

your answer is <u>D</u><u>.</u><u> </u>physical change, because a new substance is not formed.

5 0
3 years ago
Помогите пожалуйста или с первым вопросом или со вторым. Буду очень благодарна, т.к. это мой зачет... :С
Bess [88]
Hsiaqoanwbwiso iDisks
5 0
3 years ago
A sample of metal has a mass of 16.5916.59 g, and a volume of 5.815.81 mL. What is the density of this metal
Oxana [17]

Answer:

=> 2.8554 g/mL

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m)     = 16.59 g

Volume (v) =  5.81 mL

From our question, we are to determine the density (rho) of the rock.

The formula:

p = \frac{m}{v}

Substitute the values into the formula:

​p = \frac{16.59 g}{5.81 mL} \\   = 2.8554 g/mL

= 2.8554 g/mL

Therefore, the density (rho) of the rock is 2.8554 g/mL.

5 0
2 years ago
What is a specialized
ale4655 [162]

What's your question? Am I missing something?

7 0
3 years ago
Read 2 more answers
If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?
IceJOKER [234]

Answer:

n_{H_2O}=1.5molH_2O

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O

Best regards.

5 0
3 years ago
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