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4 years ago

Major species present when dissolved in water ammonium iodide

1 answer:

5 0

Ammonium iodide, or NH₄I, is a salt. Since all salts are strong electrolytes, when ammonium iodide is dissolved in water, it would dissociate into ammonium ions and iodide ions. Hence, the major species present would be: NH₄⁺ and I⁻ ions.

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Many chemistry problems result in equations of the form

Gnom [1K]

Answer:

When this equation is solved, the two values of the unknown are 0.0643 and -0.082

Explanation:

Given

$1.77 * 10^{-2} = \frac{x^2}{0.298 - x}$ --- the actual equation

Required

The values of x

We have:

$1.77 * 10^{-2} = \frac{x^2}{0.298 - x}$

Cross Multiply

$1.77 * 10^{-2} * (0.298 - x)= x^2$

Multiply both sides by 100

$1.77 * (0.298 - x)= 100x^2$

Open bracket

$0.52746 - 1.77x= 100x^2$

Rewrite as:

$100x^2 + 1.77x - 0.52746 =0$

Using quadratic formula:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where:

$a = 100; b = 1.77; c = -0.52746$

So, we have:

$x = \frac{-1.77 \± \sqrt{1.77^2 - 4*100*- 0.52746 }}{2*100}$

$x = \frac{-1.77 \± \sqrt{214.1169}}{2*100}$

$x = \frac{-1.77 \± 14.63}{200}$

Split

$x = \frac{-1.77 + 14.63}{200}\ or\ x = \frac{-1.77 - 14.63}{200}$

$x = \frac{12.86}{200}\ or\ x = \frac{-16.40}{200}$

$x = 0.0643\ or\ x = -0.082$

3 0

3 years ago

If 152 grams of ethane (c2h6) are reacted with 231 grams of oxygen gas, what is the mass of the excess reactant leftover after t

Naya [18.7K]

Answer is: the mass of the excess reactant (ethane) leftover is 90.135 grams.
Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g).
m(C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.

6 0

4 years ago

Read 2 more answers

Predict the nature of the indicated covalent bond.

11111nata11111 [884]

non-polar

Explanation :

Hydrogen and carbon have similar electronegativity values, so the C—H bond is not normally considered a polar covalent bond.

4 0

2 years ago

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago

In terms of chemical bonds and electrons, what kinds of changes occur between atoms when substances undergo chemical reactions?

STALIN [3.7K]

In a chemical change, the atoms in the reactants rearrange themselves and bond together differently to form one or more new products with different characteristics than the reactants. When a new substance is formed, the change is called a chemical change.
www.middleschoolchemistry.com/lessonplans/chapter6/lesson1

8 0

4 years ago