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vichka [17]
3 years ago
12

how many kilograms of water must evaporate from 8kg of a 25% salt solution to produce 40% salt solution?

Chemistry
1 answer:
Lena [83]3 years ago
6 0

Answer: The kilograms of water must evaporate from 8kg of a 25% salt solution to produce 40% salt solution is 3 kg.

Explanation:

According to the ratio and proportion:

C_1m_1=C_2m_2

where,

C_1 = concentration of ist solution = 25%

m_1 = mass of ist solution = 8 kg

C_2 = concentration of second solution = 40%

m_2 = mass of second solution = ? kg

25\times 8=40\times m_2

m_2=5kg

Thus the final solution must have a mass of 5 kg , i.e (8-5)= 3 kg of mass must be evaporated.

Therefore, the mass that must be evaporated from 8kg of a 25% salt solution to produce 40% salt solution is 3 kg.

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8 0
3 years ago
Increasing the intensity and duration of an exercise program too quickly can lead to injury.
Maslowich
<span>It is true that a sudden increase in both intensity and duration of physical exercise can result in injury. The body needs time to process any changes. Muscles need to build strength and flexibility gradually and the joints need time to acclimate to changes in pressure.</span>
6 0
3 years ago
Read 2 more answers
If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T
zhuklara [117]

Answer: The actual yield of B_2O_3 is 60.0 g

Explanation:-

The balanced chemical reaction :

B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)

Mass of B_2H_6 =Density\times Volume=1.131g/ml\times 36.9ml=41.7g

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles

According to stoichiometry:

1 mole of B_2H_6 gives = 1 mole of B_2O_3

1.51 moles of B_2H_6 gives =\frac{1}{1}\times 1.51=1.51 moles of B_2O_3

Theoretical yield of B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g

Percent yield of B_2O_3= 75.7\%

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

75.7\%=\frac{\text{Actual yield}}{79.3}\times 100

{\text{Actual yield}}=60.0g

Thus the actual yield of B_2O_3 is 60.0 g

7 0
3 years ago
An isotope of Cesium -137 has a half-life of 30 years. If 1.0 g of Cesium – 137 disintegrates over a period of 60 years, how man
Nady [450]

Answer:

Explanation:

Radioactive decay follows the equation:

Ln [A] = -kt + ln [A]₀

<em>Where [A] is amount of isotope after time t: Our incognite,</em>

<em>k is rate constant: ln 2 / Half-life = 0.0231 years⁻¹</em>

<em>t are 60 years</em>

<em>[A]₀ is initial amount of isotope: 1.0g</em>

<em />

Replacing:

Ln [A] = -kt + ln [A]₀

Ln [A] = -0.0231 years⁻¹*60 years + ln 1.0g

ln [A] = -1.386

[A] = 0.25g

<em />

4 0
3 years ago
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