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Lorico [155]
2 years ago
8

How many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0g of chlorine? what is the limiting reactant?

Chemistry
1 answer:
zloy xaker [14]2 years ago
4 0

Answer:

m_{HCl}=36.1gHCl

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2

Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl

Regards!

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Answer:

empirical formula = C3H7

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3 0
3 years ago
A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o
HACTEHA [7]

Answer:

1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Explanation:

From the question given above, the following were obtained:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Volume of solution = 20.4 mL

Concentration of solution =?

Next, we shall determine the mass of sodium hydroxide, NaOH. This can be obtained as as illustrated below:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Mass of NaOH =?

Mass of NaOH = (Mass Cylinder + NaOH) – (Mass of empty cylinder)

Mass of NaOH = 31.92 – 9.5

Mass of NaOH = 22.42 g

Finally, we shall determine concentration of the solution as follow:

Mass of NaOH = 22.42 g

Volume of solution = 20.4 mL

Concentration of solution =?

Concentration = mass /volume

Concentration of solution = 22.42 / 20.4

Concentration of solution = 1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Therefore, the concentration of the solution is 1.1 gmL¯¹

3 0
3 years ago
What does Gibbs free energy depend on? O A. It depends on the rate and quantity of the reaction. O B. It depends on the activati
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Answer:

C: It depends on the entropy and enthalpy of the reaction.

Explanation:

Gibbs free energy is defined as the maximum amount of non-expansion work that can be gotten from a closed system. Now this work is usually done in place of the system’s internal energy and Energy that is not extracted as work is usually exchanged with the immediate surroundings in the form of heat.

5 0
3 years ago
From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
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<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

7 0
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Answer: The momentum of the child and milk together is 58.125 kg.m/s

Explanation:

Momentum is defined as the product of object's mass and velocity.

Mathematically,

p=mv

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m = mass of the object

v = velocity of the object

In the given question, we are given that a child of mass 21.0 kg is carrying a gallon of milk having mass 2.25 kg and running with a velocity of 2.5 m/s. Hence, the momentum by both milk and child will be:

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Given:

m_{child}=21.0kg\\m_{milk}=2.25kg\\v=2.5m/s\\p=?kg.m/s

Putting values in equation 1, we get:

p=(21+2.25)kg\times 2.5m/s=58.125kg.m/s

Hence, the momentum of the child and milk together is 58.125 kg.m/s

7 0
3 years ago
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