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4 years ago

PLEASE HELP ME!!! ASAP

Chemistry

1 answer:

shtirl [24]4 years ago

5 0

Answer:

Theoretical yield of the reaction = 34 g

Excess reactant is hydrogen

Limiting reactant is nitrogen

Explanation:

Given there is 100 g of nitrogen and 100 g of hydrogen

Number of moles of nitrogen = 100 ÷ 28 = 3·57

Number of moles of hydrogen = 100 ÷ 2 = 50

Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation

N2 + 3H2 → 2NH3

From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed

Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen

⇒ We require 10·71 moles of hydrogen

But we have 50 moles of hydrogen

∴ Limiting reactant is nitrogen and excess reactant is hydrogen

From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia

Molecular weight of ammonia = 17 g

∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g

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Answer:

To supply the required ions it is necessary to inject 5,6mL of 6g/30mL solution and 131,1 mL of 0,9% solution.

Explanation:

1mEq of sodium are 59mg of NaCl and 1mEq of potassium are 75mg KCl

in intravenous infusion 15 mEq of K are:

15x75mg KCl = 1,125g of KCl

And 20 mEq of Na are:

20x59mg NaCl = 1,18g of NaCl

To supply the potassium ion it is necessary to inject:

1,125g of KCl× $\frac{30mL}{6g}$ = 5,6mL of 6g/30mL solution

And, to supply the sodium ion it is necessary to inject:

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I hope it helps!

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3 years ago

Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi

marta [7]

Answer : The value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

Explanation :

The following equilibrium reactions are :

(1) $2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2$ $K_1=5.40\times 10^{-16}$

(2) $2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)$ $K_2=1.06\times 10^{10}$

(3) $CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)$ $K_3=2.68\times 10^{-9}$

The final equilibrium reaction is :

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$ $K_{goal}=?$

Now we have to calculate the value of $K_{goal}$ for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{goal}=\sqrt{K_1\times K_2\times K_3}$

Now put all the given values in this expression, we get :

$K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}$

$K_{goal}=1.238\times 10^{-7}$

Therefore, the value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

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3 years ago

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valina [46]

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calculation

concentration = moles /volume in liters

volume in liters = 150 /1000= 0.15 L

number of moles calculation

write the equation for dissociation of Al2Cl3

that is AlCl3 ⇔ Al^3+ + 3 Cl ^-

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