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vredina [299]
3 years ago
12

I really need help guys in order to pass this. its kinda urgent!

Chemistry
1 answer:
lbvjy [14]3 years ago
6 0
Female energy the answer is the first one
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Is calcium carbide ionic or covalent compound ?
Anna007 [38]
Calcium carbide is a covalent compound. 
5 0
3 years ago
Read 2 more answers
The density of a rock that was a mass of 454 grams and a volume of 100 cm3
puteri [66]

Answer:

The answer is

<h2>4.54 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass of rock = 454 g

volume = 100 cm³

The density of the rock is

density =  \frac{454}{100}  =  \frac{227}{50}  \\

We have the final answer as

<h3>4.54 g/cm³</h3>

Hope this helps you

3 0
3 years ago
In a reaction process a company expected to make 6.8
svetlana [45]

Answer:

1) 61.8%

Explanation:

(4.2/6.8)*100

3 0
3 years ago
What U.S. state is directly north of Cuba?
Arturiano [62]

Answer:

Florida

Explanation:

Directly north of Cuba is the rural bar fight of a state, Florida

6 0
3 years ago
Consider the following reaction. 3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g) At 900°C, Kc for the reaction is 5.1. If 0.050 mol of H2O
Oduvanchick [21]

Answer:

m_{Fe_3O_4}=1.7gFe_3O_4

Explanation:

Hello,

In this case, considering the given reaction:

3 Fe(s) + 4 H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4 H_2(g)

Thus, for the equilibrium, just water and hydrogen participate as iron and iron(II,III) oxide are solid:

Kc=\frac{[H_2]^4}{[H_2O]^4}

Thus, at the beginning, the concentration of water is 0.05 M and consequently, at equilibrium, considering the ICE procedure, we have:

5.1=\frac{(4x)^4}{(0.05M-4x)^4}

Thus, the change x is obtained as:

\sqrt[4]{5.1} =\sqrt[4]{[\frac{(4x)}{(0.05M-4x)}]^4}\\\\1.5=\frac{(4x)}{(0.05M-4x)}\\\\x=0.0075M

Thus, the moles of hydrogen at equilibrium are:

[H_2]_{eq}=4*0.0075\frac{mol}{L}*1.0L=0.03molH_2

Therefore, the grams of iron(II,III) oxide finally result:

m_{Fe_3O_4}=0.03molH_2*\frac{1molFe_3O_4}{4molH_2}*\frac{231.533gFe_3O_4}{1molFe_3O_4} \\m_{Fe_3O_4}=1.7gFe_3O_4

Best regards.

6 0
3 years ago
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