How many more valence electrons does this atom need to fulfill the Octet Rule?

Suggest two observations that would be made when rubidium is added to cold water.

tino4ka555 [31]

Answer:

1. Rubidium metal reacts very rapidly with water to form a colorless basic solution of rubidium hydroxide (RbOH) and hydrogen gas (H2).

2. Rubidium sinks because it is less dense than water. It reacts violently and immediately, with everything leaving the container. Rubidium hydroxide solution and hydrogen are formed.

6 0

3 years ago

Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in

IgorC [24]

Answer:

669.48 kJ

Explanation:

According to the question, we are required to determine the heat change involved.

We know that, heat change is given by the formula;

Heat change = Mass × change in temperature × Specific heat

In this case;

Change in temperature = Final temp - initial temp

= 99.7°C - 20°C

= 79.7° C

Mass of water is 2000 g ( 2000 mL × 1 g/mL)

Specific heat of water is 4.2 J/g°C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

= 669,480 joules

But, 1 kJ = 1000 J

Therefore, heat change is 669.48 kJ

7 0

3 years ago

Automotive air bags inflate when sodium azide, NaN3, decomposes explosively to its constituent elements. How many moles of nitro

attashe74 [19]

Answer:

2.29 g of N2

Explanation:

We have to start with the chemical reaction:

$NaN_3~->~Na~+~N_2$

The next step is to balance the reaction:

$2NaN_3~->~2Na~+~3N_2$

We can continue with the mol calculation using the molar mass of

$NaN_3$ (65 g/mol), so:

$3.55~g~NaN_3\frac{1~mol~NaN_3}{65~g~NaN_3}=0.054~mol~NaN_3$

Now, with the molar ratio between $NaN_3$ and $N_2$ we can calculate the moles of $N_2$ (2:3), so:

$0.054~mol~NaN_3\frac{3~mol~N_2}{2~mol~NaN_3}=0.0819~mol~N_2$

With the molar mass of $N_2$ we can calculate the grams:

$0.0819~mol~N_2=\frac{1~mol~N_2}{28~g~N_2}=2.29~g~N_2$

I hope it helps!

5 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

3 years ago

When you have finished adding the necessary amount of NaOH to your beaker, what color will the litmus paper turn?

Svetllana [295]

Answer: The red litmus paper turns blue on dipping in NaOH solution.

Explanation:

Litmus paper is the indicator that detects the nature of the solution, whether it is acidic or basic.

There are 2 types of litmus paper:

Red litmus paper: This paper will turn blue if it is dipped in basic solution and will remain as such if it is dipped in acidic solution.
Blue litmus paper: This paper will turn red if it is dipped in acidic solution and will remain as such if it is dipped in basic solution.

NaOH is a strong base, so when a red litmus paper is dipped in the beaker having necessary amount of NaOH, the red litmus paper turns into blue.

7 0

4 years ago