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dalvyx [7]
4 years ago
15

Do you think sodium carbonate is an acid or base?

Chemistry
1 answer:
Reil [10]4 years ago
7 0
I just looked it up and it says "Na2CO3 (sodium carbonate) is neither an acid nor a base. It is a salt..." I really hoped this helped you, if not I'm sorry.
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sesenic [268]
The answer is b, corporation
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2 years ago
The Density of this log is 1.75g/cm3. If we cut this log in half, what is the density of the log? Make sure to show your work an
swat32

Answer:

The density remains the same

Explanation:

The density remains the same because cutting the object in half will divide the mass & volume by the same amount. The density cant be changed no matter what happens to it.

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3 years ago
a rigid container holds a gas at a pressure of 0.54 atm and a temperature of -100.0c. what will the pressure be when the tempera
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Answer:

150kpa

Explanation:

using Gay-Lusscass law

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If 1.00 mol of argon is placed in a 0.500-l container at 22.0 ?c , what is the difference between the ideal pressure (as predict
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Calculate the pressure using the Van der Waals equation and the pressure using the ideal gas equation PV=nRT. Subtract the two pressures to get the difference. then:<span>Calculate how many moles of ammonia you have using the ideal gas equation PV=nRT. Multiply the number of moles by the molar mass of ammonia to get the mass in grams.</span>
5 0
3 years ago
A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o
statuscvo [17]

<u>Answer:</u> The value of K_{eq} is 0.044

<u>Explanation:</u>

We are given:

Initial moles of methane = 4.10\times 10^{-2}mol=0.0410moles

Initial moles of carbon tetrachloride = 6.51\times 10^{-2}mol=0.0651moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of methane = \frac{0.0410}{1.00}=0.0410M

Concentration of carbon tetrachloride = \frac{0.0651}{1.00}=0.0651M

The given chemical equation follows:

                    CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)

<u>Initial:</u>          0.0410    0.0651

<u>At eqllm:</u>     0.0410-x   0.0651-x       2x

We are given:

Equilibrium concentration of carbon tetrachloride = 6.02\times 10^{-2}M=0.0602M

Evaluating the value of 'x', we get:

\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M

Now, equilibrium concentration of methane = 0.0410-x=[0.0410-0.0049]=0.0361M

Equilibrium concentration of CH_2Cl_2=2x=[2\times 0.0049]=0.0098M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}

Putting values in above expression, we get:

K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044

Hence, the value of K_{eq} is 0.044

8 0
3 years ago
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