CaC2 + 2H2O → C2H2 + Ca(OH)2 If 2.8 moles of CaC2 are consumed in this reaction, how many grams of H2O are needed?

Exercise while in space is important because it PREVENTS what?

dsp73

Answer:

bone loss and muscle deterioration

Explanation:

this helps the body keep thriving and bones strong and muscles in tact

8 0

3 years ago

It is difficult to measure the volume of a gas because: Gases have no volume. Gases are not soluble in liquids. The volume of a

NeTakaya

The volume of the gas depends on the container

7 0

3 years ago

Read 2 more answers

Give the net ionic equation for the reaction (if any) that occurs when aqueous solutions of Al(C2H3O2)3 and LiNO3 are mixed. Giv

RSB [31]

Answer: No reaction occurs

Explanation:.

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The given chemical equation is:

$Al(C_2H_3O_2)_3(aq)+3LiNO_3(aq)\rightarrow Al(NO_3)_3(aq)+3LiC_2H_3O_2(s)$

The complete ionic equation is;

$Al^{3+}(aq)+3C_2H_3O_2^-(aq)+3Li^+(aq)+3NO_3^-(aq)\rightarrow Al^{3+}(aq)+3NO_3^-(aq)+3Li^+(aq)+3C_2H_3O_2^-(aq)$

The ions which are present on both the sides of the equation are are not involved in net ionic equation.

Hence, there is no net reaction.

5 0

3 years ago

In the third period of the periodic table sodium is followed by magnesium aluminum silicon and phosphorus which of these element

wariber [46]

Answer:
Phosphorous has the smallest atomic size.

Explanation:
As we know these elements belong to same period means there valence shell is the same. So moving from left to right along the period the shell number remains constant but the number of protons and electrons increases. So, due to increase in number of protons the nuclear charge increases hence attracts the valence electrons more effectively resulting in the decrease of atomic size.

Elements and their atomic radius are as follow,

Magnesium 0.160 nm

Aluminium 0.130 nm

Silicon 0.118 nm

Phosphorus 0.110 nm

6 0

3 years ago

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

3 0

2 years ago