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2 years ago

What reaction will take place if h2o is added to a mixture of nanh2/nh3? draw the products of the reaction (without sodium ion)?

Chemistry

1 answer:

Maru [420]2 years ago

8 0

When sodium amide i.e. $NaNH_{2}$ reacts with water i.e. $H_{2}O$ results in the formation of sodium hydroxide i.e. $NaOH$ and ammonia $NH_{3}$ .

The chemical reaction is given by:

$NaNH_{2}+H_{2}O\rightarrow NH_{3}+NaOH$

Now, when ammonia i.e. $NH_{3}$ reacts with water results in the formation of ammonium hydroxide i.e. $NH_{4}OH$

The chemical reaction is given by:

$NH_{3}+H_{2}O\rightarrow NH_4OH$

Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).

The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.

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Activity 7: Crossword Puzzle

Ivenika [448]

Answer:

Across

2. Conduction.

3. Plates

4. Convection

5. Subduction

7. Earthquake

Down

1. Radioactive

6. Radiation

8. Sink

9. Slabpull

The clues are;

Across:

2. air molecules come in contact with warmer molecules

3. crust are made up of puzzle - like landmass called_____

4. rising and falling movement of material in the mantle

5. when tectonic plates push with each other

7. it is the result of movement of earth's plate

Down:

1. elements that play a vital role in Earth's internal heat

6. least important mode of heat transport

8. warm material rise; cool material______

9. heats build up underneath the crust

4 0

1 year ago

12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68

luda_lava [24]

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of $H_2O$ = 1.00 kg = 1000 g

Moles of $CsF$ = 3.68 mole

Molar mass of $H_2O$ = 18 g/mole

Vapor pressure of water = 0.692 atm

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

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole$

Now we have to calculate the mole fraction of $H_2O$

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938$

Now we have to partial pressure of solution.

According to the Raoult's law,

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

where,

$P_{Solution}$ = vapor pressure of solution

$P^o_{H_2O}$ = vapor pressure of water = 0.692 atm

$X_{H_2O}$ = mole fraction of water = 0.938

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

$P_{Solution}=0.938\times 0.692atm$

$P_{Solution}=0.649atm$

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

5 0

1 year ago

The radius of a potassium atom is 231 pm. How many potassium atoms would have to be laid side by side to span a distance of 4.77

Softa [21]

To determine the number of potassium laid side by side by a given distance, we simply divide the total distance to the diameter of each atom. The diameter is twice the radius of the atom. We calculate as follows:

number of atoms = 4770 / 231x10^-12 = 2.06x10^13 atoms

3 0

2 years ago

Read 2 more answers

How many grams of the excess reagent are left over when 6.00g of CS2 gas react with 10.0g of Cl2 gas in the following reaction:

WARRIOR [948]

Answer:

CS₂ = 2.43 g

Explanation:

Data Given:

CS₂ gas = 6.00g

Cl₂ = 10.0g

Reaction Given:

CS₂(g) + 3Cl₂(g) --------> CCl₄(l) + S₂Cl₂(l)

Solution

Limiting Reagent :

The reactant which is less in amount control the amount of product and know as limiting reagent.

Excess Reagent:

The amount of reactant which are in excess and leftover at the end of reaction and product formed.

Now we have to find the reactant that is in excess

For this we will look at the Reaction

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

1 mol 3 mol 1 mol 1 mol

we come to know from the above reaction that

1 mole of CS₂ react with 3 mole of Cl₂ to produce 1 mole of CCl₄ and 1 mole of S₂Cl₂

Now to convert mole to mass we required molar masses

molar mass of CS₂ = 12 + 2(32)

molar mass of CS₂ = 76 g/mol

molar mass of Cl₂ = 2(35.5)

molar mass of Cl₂ = 71 g/mol

if we represent mole in grams then

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

1 mol (76 g/mol) 3 mol (71 g/mol)

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

76 g 213 g

So,

we come to know that 76 g of CS₂ will combine with 213 g of Cl₂ to form product.

So now look for the ratio of both reactant

CS₂ : Cl₂

76 g : 213 g

1 g : 2.8 g

So, the for every one gram of CS₂ required 2.8g Cl₂

So from this details

we apply unity formula

2.8 g of Cl₂ ≅ 1 g of CS₂

10 g of Cl₂ ≅ ? g of CS₂

by doing cross multiplication

g of CS₂ = 10 x 1 / 2.8

g of CS₂ = 3.6 g

So,

10.0g of Cl₂ used 3.57 g of CS₂

It showed that 3.57 g of CS₂ used and the remaining amount of CS₂ is

Remaining amount of CS₂ = 6 - 3.57 = 2.43 g

8 0

1 year ago

A sample of a compound of Cl and O reacts with an excess of H2 to give 0.233g of HCl and 0.403g of H2O. Determine the empirical

Vsevolod [243]

Answer:

Cl₂O₇

Explanation:

For the reaction:

ClₓOₙ + H₂ → HCl + H₂O

Moles of HCl and moles of H₂O are:

HCl: 0.233g HCl ₓ (1mol / 36.46g) = 6.39x10⁻³ mol HCl

H₂O: 0.403g H₂O ₓ (1mol / 18.02g) = 2.236x10⁻² mol H₂O

As you can see, moles of HCl are equivalent to moles of Cl in the compound and moles of H₂O are equivalent to moles of O in the compound, that means:

6.39x10⁻³ mol Cl

2.236x10⁻² mol O

Empirical formula is the simplest ratio of atoms presents in a molecule. If Cl is 1, Oxygen will be:

2.236x10⁻² mol / 6.39x10⁻³ = 3.5

As empirical formula must be given in natural numbers, the empirical formula is:

Cl₂O₇

4 0

2 years ago

Read 2 more answers