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3 years ago

When chalk is added to water and stirred what is formed

Chemistry

2 answers:

sukhopar [10]3 years ago

7 0

The chalk will most definitely dissolve after time

lianna [129]3 years ago

5 0

The chalk will dissolve

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A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coefficient is "1". KOH + Cu(NO3)2 → KNO

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2KOH + Cu(NO3)2 → 2KNO3 + Cu(OH)2

2K⁺ 1Cu²⁺ 2K⁺ 1Cu²⁺

2OH ⁻ 2NO3⁻ 2NO3⁻ 2OH⁻

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3 years ago

0.5 gm of mixture of NH4Cl and NaCl was boiled with 25 ml of 0.95 N NaOH in a vessel till all the ammonia is expelled.The residu

GalinKa [24]

Answer:

72.66%

Explanation:

NH₄Cl reacts in presence of NaOH producing ammonia, NH₃, as follows:

NH₄Cl + NaOH → NaCl + NH₃ + H₂O

The residual NaOH reacts with H₂SO₄ as follows:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

The equivalent-gram of H₂SO₄ are:

16mL * 0.1N * 1.06 = 1.696mEq.

As the complete residual solution is 100mL but the neutralization was made only with 10mL, the mEq you need to neutralize the residual NaOH is:

1.696mEq * (100mL / 10mL) = 16.96mEq.

The mEq of NaOH you add in the first are:

25mL * 0.95mEq = 23.75mEq

That means the NaOH that reacts = moles of NH₄Cl is:

23.75mEq - 16.96mEq = 6.79mEq = 6.79mmoles NH₄Cl =

6.79x10⁻³ moles NH₄Cl

In grams (Using molar mass NH₄Cl = 53.5g/mol):

6.79x10⁻³ moles NH₄Cl * (53.5g / mol) =

0.3633g of NH₄Cl are in the original mixture.

% is:

0.3633g/ 0.5g * 100 = 72.66%

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3 years ago

The two electrons in a helium atom occupy

IRINA_888 [86]

Answer:

the answer is A

Explanation:

4 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

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3 years ago

6. We have many vestigial structures that are similar to structures on modern day chimpanzees. How can we

ollegr [7]

Answer:

pineal body is a vastigious structure which signifies same origin of chimpanzee and modern man

Explanation:

5 0

3 years ago

When chalk is added to water and stirred what is formed​

When chalk is added to water and stirred what is formed