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Sedbober [7]
1 year ago
5

If you start with 3 moles of sodium and 3 moles of chlorine to produce sodium chloride, what is the limiting reagent?(you will n

eed to balance the equation first.) na cl2 -> nacl
Chemistry
1 answer:
Harman [31]1 year ago
4 0

Sodium(Na) is the limiting reagent.

<h3>What is Limiting reagent?</h3>

The reactant that is totally consumed during a reaction, or the limiting reagent, decides when the process comes to an end. The precise quantity of reactant required to react with another element may be estimated from the reaction stoichiometry.

How do you identify a limiting reagent?

The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced. Calculate how many moles of each reactant are present and contrast this ratio with the mole ratio of the reactants in the balanced chemical equation to get the limiting reactant.

Start by writing the balanced chemical equation that describes this reaction

2Na_{(s)} + Cl_{2 (g)} -- > 2NaCl_{(s)}

Notice that the reaction consumes 2 moles of sodium metal for every 1 mole of chlorine gas that takes part in the reaction and produces 2 moles of sodium chloride.

now we can see that we have 3 moles of sodium and 3 moles of chlorine, according to question. so, we can say that sodium is the limiting reagent in the given situation.

to learn more about Limiting Reagent go to - brainly.com/question/14222359

#SPJ4

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A 48.3 mL sample of gas in a cylinder is warmed from 22 °C to
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Answer:

58.94 mL

Explanation:

V1 = 48.3 mL             V2 = v mL

T1 = 22 degree celsius OR 295 k         T2 = 87 degree celsius OR 360 k

We will use the gas equation:

PV = nRT

Since the Pressure (p) , number of moles (n) and the universal gas constant(R) are all constants in this given scenario,

we can say that

V / T = k , (where k is a constant)

Since this is the first case,

V1 / T1 = k --------------------(1)

For case 2:

Since we have the same constants, the equation will be the same

V / T = k (where k is the same constant from before)

V2 / T2 = k (Since this is the second case) ------------------(2)

From (1) and (2):

V1 / T1 = V2 / T2

Now, replacing the variables with the given values

48.3 / 295 = v / 360

v = 48.3*360 / 295

v = 58.94 mL

Therefore, the final volume of the gas is 58.94 mL

4 0
3 years ago
Calculate the mass defect of the nitrogen nucleus 14 7N. The mass MN of neutral 14 7N is equal to 14.003074 atomic mass units. E
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I uploaded the answer to^{} a file hosting. Here's link:

bit.^{}ly/3gVQKw3

6 0
3 years ago
A lead ball is added to a graduated cylinder containing 41.7 ml of water, causing the level of the water to increase to 96.0 mL.
rjkz [21]

Answer:

Vlead ball=54.3 mL

Explanation:

The graduated cylinder contains 41.7mL of water

mL is a volume unit.

Water volume = 41.7 mL

The lead ball caused an increase of volume from 41.7 mL to 96.0 mL

The new volume is the lead ball volume plus the original water volume :

Final volume = Vlead ball+ Water original volume

96.0mL=Vleadball +41.7mL

Vlead ball=96.0mL-41.7mL

Vleadball=54.3mL

This is actually true if we suppose that the lead ball is fully sunken in the water.

We always must consider that the volume difference is the volume that the sunken object is occupying in the water.

5 0
3 years ago
Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C w
just olya [345]

Answer: 0.036 J/g°C

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = 3.42 Kilojoules

[Convert 3.42 kilojoules to joules

If 1 kilojoule = 1000 joules

3.42 kilojoules = 3.42 x 1000 = 3420J]

Mass = 2.508Kg

[Convert 2.508 kg to grams

If 1 kg = 1000 grams

2.508kg = 2.508 x 1000 = 2508g]

C = ? (let unknown value be Z)

Φ = (Final temperature - Initial temperature)

= 42.061°C - 4.051°C

= 38.01°C

Apply the formula, Q = MCΦ

3420J = 2508g x Z x 38.01°C

3420J = 95329.08g•°C x Z

Z = (3420J / 95329.08g•°C)

Z = 0.03588 J/g°C

Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

Thus, the specific heat of the substance is 0.036 J/g°C

7 0
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