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3 years ago

How many grams of NaCl should be obtained to make 150 mL of 4.5 M solution

Chemistry

1 answer:

nadya68 [22]3 years ago

3 0

Answer:

39,5 grams should be obtained of NaCl

Explanation:

We calculate the weight of 1 mol of NaCl:

Weight 1 mol NaCl= Weight Na + Weight Cl= 23g + 35,5g=58,5 g/mol

4,5M--> 4,5 moles NaCl in 1000ml (1L) of solution

1000ml-----4,5 moles NaCl

150 ml------x=(150 mlx4,5 moles NaCl)/1000ml=0,675 moles NaCl

1 mol NaCl--------------58,5 grams

0,675molesNaCl---x= (0,675molesNaClx58,5 grams)/1 mol NaCl

x= 39,4879 grams

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How many moles of oxygen are required to produce 37.15 g CO2?

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Carbon = 12.010. Oxygen = 15.999 x 2 15.999 x 2 = 31.998 + 12.010 = 44.008 \frac{37.15 grams * 1 mole CO2}{44.008 grams}

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The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0°c is 4.48. what is the value of ka for hbro?

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The solution would be like this for this specific problem:

Given:

pH of a 0.55 M hypobromous acid (HBrO) at 25.0 °C = 4.48

[H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-] 

Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9

To add, Hypobromous Acid does not require acid adjustment, which is necessary for chlorine-based product and is stable and effective in pH ranges of 5-9.

Hypobromous Acid combines with organic compounds to form a bromamine. Chlorine also combines with the same organic compounds to form a chloramine. It is also one of the least expensive intervention antimicrobial compounds available.

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3 years ago

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An electrically neutral atom bears what trait?

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C) It contains the same number of electrons and protons.

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4 years ago

One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su

Tpy6a [65]

Answer : The percent purity of $Fe_2O_3$ in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

$Fe_2O3+3CO\rightarrow 2Fe+3CO_2$

First we have to calculate the mass of Fe.

$\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}$

Molar mass of Fe = 55.8 g/mole

$\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole$

Now we have to calculate the moles of $Fe_2O_3$

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of $Fe_2O_3$

So, $3.15\times 10^4mole$ of Fe produced from $\frac{3.15\times 10^4}{2}=15750$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3$

Molar mass of $Fe_2O_3$ = 159.69 g/mole

$\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg$

Now we have to calculate the percent purity of $Fe_2O_3$ in the original sample.

Mass of original sample = $2.86\times 10^3kg$

$\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100$

$\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%$

Therefore, the percent purity of $Fe_2O_3$ in the original sample is 87.94 %

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Answer:

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Explanation:

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