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3 years ago

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The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abb

reviation (w/v) is also common. How many grams of sucrose are needed to make 725 mL of a 35.0% (w/v) sucrose solution?

1 answer:

Nimfa-mama [501]3 years ago

3 0

Answer:

$\boxed{\text{254 g}}$

Explanation:

$\begin{array}{rcl}\text{\% m/V} & = & \dfrac{\text{Mass of sucrose}}{\text{Volume of solution}}\\\\\text{Let m}& = &\text{mass of sucrose}\\\dfrac{\text{35.0 g}}{\text{100 mL}}& = & \dfrac{m}{\text{725 mL}}\\\\m & = &\dfrac{\text{35.0 g}\times 725}{100}\\\\ & = &\textbf{254 g}\\\end{array}\\\text{You need $\boxed{\textbf{254 g}}$ of sucrose}$

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4.07 x 10-17<br> (5.6 x 10") (5.8 x 105)

Andrej [43]

4.07 x 10-17 = 23.7
(5.6 x 10”) (5.8 x 105)= log(100)
log(7)
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3 0

3 years ago

How many moles of ammonium ions are in 125 mL of 1.40 M NH4NO3 solution? ________ moles (give answer with correct sig figs in un

Sholpan [36]

The number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

We'll begin by calculating the number of mole of NH₄NO₃ in the solution. This can be obtained as follow:

Volume = 125 mL = 125 / 1000 = 0.125 L

Molarity = 1.40 M

<h3>Mole of NH₄NO₃ =? </h3>

Mole = Molarity x Volume

Mole of NH₄NO₃ = 1.40 × 0.125

<h3>Mole of NH₄NO₃ = 0.175 mole</h3>

Finally, we shall determine the number of mole of ammonium ion, NH₄⁺ in the solution. This can be obtained as follow:

NH₄NO₃(aq) —> NH₄⁺(aq) + NO₃¯(aq)

From the balanced equation above,

1 mole of NH₄NO₃ contains 1 mole of NH₄⁺

Therefore,

0.175 mole of NH₄NO₃ will also contain 0.175 mole of NH₄⁺

Thus, the number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

Learn more: brainly.com/question/25469095

3 0

2 years ago

What does the cell theory state?

strojnjashka [21]

C, that cells cannot be created nor destroyed.

Cells are produce by other cells. Cells can not be man-made nor destroyed.

6 0

3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

Calculate the grams of CaCl2 necessary to make a 0.15Msolution.

xxTIMURxx [149]

Answer: mass m = M·c·V

Explanation: M(CaCl2) = 110.98 g/mol, c= 0.15 mol/l,

n=m/M= cV, volume of Solution is not mentioned

3 0

2 years ago