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4 years ago

What does the symbol Cp stand for?

Chemistry

2 answers:

notka56 [123]4 years ago

7 0

It should be Copernicium

nikklg [1K]4 years ago

3 0

Coperinicium, I believe.

Best of luck!

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Answer: Magnets have a force which is known as no contact force. This force tries to attract the metallic objects near it, which makes it move.

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3 years ago

What is the function of the nervous system? What is the basic unit of the nervous system?

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2 years ago

Read 2 more answers

Calculate δs∘rxn for the reaction2no(g) o2(g)→2no2(g)express your answer to one decimal place and include the appropriate units

Anna35 [415]

The δs∘rxn for the reaction $2NO(g) + O_{2}$ → $2NO_{2} (g)$ will be -146 J/K.

Entropy would be a measurable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty.

Entropy would be a measurement of the system's unpredictability or disorder. The entropy increases as randomness do. It has broad properties as well as a state function. It has the unit $JK^{-1} mol^{-1}$ .

Entropy of the reaction can be calculated by the reaction.

Δ $S^{0} rxn$ = 2 mol × $S^{0} (NO_{2} (g) - 2 mol$ × $S^{0} NO (g)$ - 1 mol × $S^{0} (O_{2} )$

Δ $S^{0} rxn$ = 2 mol × 240 J/mol.K - 2 mol × 210 J/mol.K-1 mol ×205.2 J/mol.K

Δ $S^{0} rxn$ = -146.8 J/K

Therefore, the δs∘rxn for the reaction $2NO(g) + O_{2}$ → $2NO_{2} (g)$ will be -146 J/K.

To know more about reaction

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4 0

1 year ago

Can someone plz help me with this

lara31 [8.8K]

A would be states of matter

B would be density

C would be Liquid

J is oxidation

H would be acid

D would be gas

I would be base

thats all i remember

7 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

4 years ago