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3 years ago

In the formation of alum, the initial filtration step is used to

Chemistry

1 answer:

Goryan [66]3 years ago

6 0

Answer:

Explanation:

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I really need help with this problem.

Aleksandr [31]

Answer:

.371 mole of NaCl

Explanation:

Na Cl Mole weight = 22.989 + 35.45 = 58.439 g/mole

21.7 g / 58.439 g/mole = .371 mole

8 0

2 years ago

Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Ac

olga2289 [7]

Answer:

1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrochloric acid = n

Volume of hydrochloric acid solution = 200.0 mL = 0.200 L

Molarity of the hydrochloric acid = 0.089 M

$n=0.089 M\times 0.200 L=0.0178 mol$ of HCL

$HCl(aq)+NaHCO_3(aq)\rightarrow NaCl(aq)+H_2O(l)+CO_2(g)$

According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.

Then 0.0178 moles of HCl wil be neutralized by :

$\frac{1}{1}\times 0.0178 mol=0.0178 mol$ of sodium bicarbonate

Mass of 0.0178 moles of sodium bicarbonate:

0.0178 mol × 72 g/mol = 1.4952 g

1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.

8 0

3 years ago

An ionic bond forms between sodium chorine when the sodium atoms _______ electrons.

Natalija [7]

is 7 cus the valance electron for sodium is Ne 3s1 and for chlorine is Ne 3s13p5

4 0

3 years ago

What is the concentration of a 1:10 dilution of a 2.5 M solution of NaCl? How would you prepare exactly 100 ml of such a solutio

Marina CMI [18]

Answer: The concentration of the solution is 0.25 M

Explanation:

Let the volume of solution of 2.5 M NaCl be 10 mL

We are given:

Dilution ratio = 1 : 10

So, the solution prepared will have a volume of = $\frac{10}{1}\times 1000=100mL$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated NaCl solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted NaCl solution

We are given:

$M_1=2.5M\\V_1=10mL\\M_2=?M\\V_2=100mL$

Putting values in above equation, we get:

$2.5\times 10=M_2\times 100\\\\M_2=\frac{2.5\times 10}{100}=0.25M$

Hence, the concentration of the solution is 0.25 M

6 0

3 years ago

Balance the chemical equation: KOH + H3AsO4 → K2HAsO4 + H2O

Colt1911 [192]

2 KOH +1 H3AsO4 →1 K2HAsO4 + 2 H2O

3 0

3 years ago