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lara31 [8.8K]
3 years ago
9

A pupil adds 5 cm of 12 mol.dm sulphuric acids to make a 250 cmº solution. Calculate the

Chemistry
1 answer:
klemol [59]3 years ago
8 0

Answer:

0.500 mol/dm³

Explanation:

Using the formula below;

CaVa = CbVb

Where;

Ca = concentration of acid (mol/dm³)

Cb = concentration of base (mol/dm³)

Va = volume of acid (cm³)

Vb = volume of base (cm³)

In accordance to the information provided in this question is;

Va = 5cm³

Vb = 250 cm³

Ca = 12 mol/dm³

Cb = ?

Using CaVa = CbVb

12 × 5 = Cb × 250

60 = 120Cb

Cb = 60/120

Cb = 0.500 mol/dm³

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Create the Equation: What is the Percent Yield of Ammonia (NH3) if 11.8 g is recovered in a reaction with 7.02 x 10^23 molecules
insens350 [35]

Answer:

Explanation:

The first thing that you need to do here is to calculate the theoretical yield of the reaction, i.e. what you get if the reaction has a

100

%

yield.

The balanced chemical equation

N

2

(

g

)

+

3

H

2

(

g

)

→

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NH

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(

g

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tells you that every

1

mole of nitrogen gas that takes part in the reaction will consume

3

moles of hydrogen gas and produce

1

mole of ammonia.

In your case, you know that

1

mole of nitrogen gas reacts with

1

mole of hydrogen gas. Since you don't have enough hydrogen gas to ensure that all the moles of nitrogen gas can react

what you need

3 moles H (sub 2)

>

what you have

1 mole H (sub2)

you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of nitrogen gas will get the chance to take part in the reaction.

So, the reaction will consume

1

mole of hydrogen gas and produce

1

mole H

2

⋅

2 moles NH

3

3

moles H

2

=

0.667 moles NH

3

at

100

%

yield. This represents the reaction's theoretical yield.

Now, you know that the reaction produced

0.50

moles of ammonia. This represents the reaction's actual yield.

In order to find the percent yield, you need to figure out how many moles of ammonia are actually produced for every

100

moles of ammonia that could theoretically be produced.

You know that

0.667

moles will produce

0.50

moles, so you can say that

100

moles NH

3

.

in theory

⋅

0.50 moles NH

3

.

actual

0.667

moles NH

3

.

in theory

=

75 moles NH

3

.

actual

Therefore, you can say that the reaction has a percent yield equal to

% yield = 75%

−−−−−−−−−−−−−

or 75 moles NH sub3

I'll leave the answer rounded to two sig figs.

5 0
3 years ago
If I have 12.0 volume of gas at pressure of 1.10 and temperature of 200k, what is the number of moles? (R=8314)
arsen [322]

Answer:

7.94 x 10^6 mol

Explanation:

PV=nRT

n=(PV)/(RT)

n=(1.10*12.0)/(8314*200)

n=7.94 x 10^6 mol

5 0
3 years ago
The highly reactive elements in group 7a are known for forming salts.what are they called
yaroslaw [1]
Group 7A are halogens 

If you look at a periodic table, these elements include F, Cl, Br, I, and At. Some well known salts are KCl and NaCl (better known as table salt!)
7 0
3 years ago
Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre
weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)

which is equivalent to

\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)

The question states that the second equation has an enthalpy, or "heat", of neutralization of -56.2 \; \text{kJ}. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 56.2 \; \text{kJ} or 56.2 \times 10^{3}\; \text{J} of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J} of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or 1.00 \times 10^{3} \; \text{ml} of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 1.00 \times 10^{3} \; \text{g}.

The solution has a specific heat of 4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. The solution thus have a heat capacity of 4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of 4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 1.405 \times 10^{4} \; \text{J} are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0
3 years ago
What is the mass of 3.01x1023 atoms of iron(atomic mass of fe=56)
nlexa [21]

Answer:

N = n× l

N = number of entities

n= moles

l = Avogadro's constant = 6.023 × 10^23

3.01 × 10^ 23 = n * 6.023 × 10^23

n = 3.01 × 10^23/6.023 × 10^23

n= 0.5moles

Molar mass = mass/ number of moles

Molar mass = 56

mass = 56 × 0.5

= 28g

Hope this helps.

4 0
4 years ago
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