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Anna71 [15]
2 years ago
10

Which of the following is a characteristic science?

Chemistry
1 answer:
NikAS [45]2 years ago
6 0

Answer:

what are the answer choices?

Explanation:

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When equal moles of an acid and a base are mixed, after reaction the two are compounds are said to be at the _______________. Se
Elodia [21]

Answer:

when equal moles of an acid and base are mixed,after reaction the two are compounds are said to be at the Equivalent point.

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How to prepare magnesium carbonate starting with magnesium nitrate
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Answer:

How to prepare Magnesium Carbonate:

Explanation:

Magnesium carbonate can be prepared in laboratory by reaction between any soluble magnesium salt and sodium bicarbonate: MgCl2(aq) + 2NaHCO3(aq) → MgCO3(s) + 2NaCl(aq) + H2O(l) + CO2(g)

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3 years ago
Which of the following definitions characterizes an element?
Amanda [17]
<span>The answer is A because an elements only consists of one type of atom. Example. the element H (hydrogen) only contains 1 hydrogen atom. I hope this helps!! </span>
6 0
3 years ago
Which of the following elements has the highest electronegativity?
blsea [12.9K]

Answer:

B) Florine

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7 0
3 years ago
Read 2 more answers
Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L contai
Zarrin [17]

Answer : The concentration of HI and I_2 at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

First we have to calculate the concentration of H_2, I_2\text{ and }HI

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.00623mol}{1.50L}=0.00415M

\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=\frac{0.00414mol}{1.50L}=0.00276M

\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=\frac{0.0244mol}{1.50L}=0.0163M

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                            2HI(g)\rightleftharpoons H_2(g)+I_2(g)

Initial conc.      0.0163     0.00415      0.00276

At eqm.        (0.0163-2x) (0.00415+x)  (0.00276+x)

As we are given:

Concentration of H_2 at equilibrium = 0.00467 M

That means,

(0.00415+x) = 0.00467

x = 0.00026 M

Concentration of HI at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M

Concentration of I_2 at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

8 0
3 years ago
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