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3 years ago

15

During science lab, Jacob mixes together sodium hydroxide (NaOH) and hydrochloric acid (HCl). The two chemicals react to form so

dium chloride (NaCl) and water (H2O).

1 answer:

Illusion [34]3 years ago

6 0

HCl + NaOH ----> NaCl + H2O

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If 5.58 g of iron reacts with sulfur to produce 8.79 g of iron sulfide, what is the mass of reacting sulfur? A) 3215 B) 14:37 C)

gogolik [260]

Answer: A) 3.21 g

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side.

$Fe+S\rightarrow FeS$

We are given:

Mass of iron = 5.58 g

Mass of iron sulphide = 8.79 g

Mass of sulphur = x g

Total mass on reactant side = 5.58 + x

Total mass on product side = 8.79 g

Applying law of conservation of mass, we get:

$5.58+x=8.79\\\\x=3.21g$

Hence, the mass of reacting sulfur is 3.21 g.

7 0

3 years ago

If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams

Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of $KCl_3$ = 2.0 moles

Molar mass of $O_2$ = 32 g/mole

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2KClO_3\rightarrow 2KCl+3O_2$

From the balanced reaction we conclude that

As, 2 mole of $KClO_3$ react to give 3 mole of $O_2$

So, 2.0 moles of $KClO_3$ react to give $\frac{2.0}{2}\times 3=3.0$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g$

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}$

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = $214.0^oC=273+214.0=487K$

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

$\rho$ = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

$P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}$

$P=1.78atm$

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0

3 years ago

Which property of a body cannot be changed if a force is applied to it?

prohojiy [21]

The answer is it’s mass

8 0

3 years ago

Read 2 more answers

In the laboratory you are asked to make a 0.565 m sodium bromide solution using 315 grams of water. How many grams of sodium bro

Scorpion4ik [409]

Answer : The mass of sodium bromide added should be, 18.3 grams.

Explanation :

Molality : It is defined as the number of moles of solute present in kilograms of solvent.

Formula used :

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}$

Solute is, NaBr and solvent is, water.

Given:

Molality of NaBr = 0.565 mol/kg

Molar mass of NaBr = 103 g/mole

Mass of water = 315 g

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

$0.565mol/kg=\frac{\text{Mass of NaBr}\times 1000}{103g/mole\times 315g}$

$\text{Mass of NaBr}=18.3g$

Thus, the mass of sodium bromide added should be, 18.3 grams.

6 0

3 years ago

Which of the following is not part of the proper protocol for using acids and bases?

Anni [7]

Answer:

B.Add acid to water,not water to acid

Explanation:

they should not be mixed

8 0

3 years ago