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3 years ago

I need to find the chemical equation

Chemistry

1 answer:

Yuri [45]3 years ago

4 0

Answer:

2NaCN + CaCO3 --> Na2CO3 + Ca(CN)2

Explanation:

Knowing the names gets us: NaCN + CaCO3 --> Na2CO3 + Ca(CN)2

Balance: there are two sodiums and cyanides on the product side so add a 2 to the reactant side.

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16. Explain why antifreeze is used in vehicle radiators. Use your knowledge of the properties of

Fynjy0 [20]

Answer:

Antifreeze is whats used to keep your engine cool without freezing.

Explanation:

it keeps the engine from overheating.

It also prevents corrosion.

Here is a quote from google "Antifreeze works because the freezing and boiling points of liquids are “colligative” properties. This means they depend on the concentrations of “solutes,” or dissolved substances, in the solution. A pure solution freezes because the lower temperatures cause the molecules to slow down"

That quote is from "The Science Behind Antifreeze"

If you have any questions feel free to ask in the comments.

4 0

3 years ago

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

3 years ago

Show your work. 37.2 moles CO2 to oxygen atoms.

Rufina [12.5K]

4 moles of oxygen (6.0zzx10

4 0

3 years ago

If you mix equal concentrations of reactants and products, will the reaction proceed to the right or the left? HF(aq)+NO3−(aq)⇌H

kaheart [24]

Answer:

hypompast

Explanation:

4 0

3 years ago

Which of the following terms describes an accumulation of rocky, sandy, or clayey material deposited at the end of a glacier?

OLEGan [10]

Answer:

Option (D)

Explanation:

Eskers are the long ridges that are comprised of rocks, sands and clay particles and are deposited towards the end of the glaciers. These are fluvioglacial depositional features. These particles are exposed after the glaciers recede. These ridges are formed parallel to the earlier flow direction of ice. The size of eskers is generally smaller as it carries smaller particles such as rocks, sands, and gravels, in comparison to the different type of moraines. It is because the flow velocity decreases as the glaciers melt. So, these eskers are formed at the end of the glaciers.

Thus, the correct answer is option (D).

3 0

3 years ago