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Alex17521 [72]
2 years ago
12

46.0 g of barium nitrate is dissolved in 360 g of water. What is the percent concentration of the barium nitrate solute in the s

olution?
Pls help with answer I don’t get it
Chemistry
1 answer:
avanturin [10]2 years ago
8 0

The percent concentration of the Barium nitrate : 11.33%

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

\tt \%mass=\dfrac{mass~solute}{mass~solution}\times 100\%

mass solute=mass Barium nitrate(Ba(NO₃)₂)=46 g

mass solvent=mass water=360 g

mass solution=mass solute+mass solvent=46+360=406 g

\tt \%mass=\dfrac{46}{406}\times 100\%\\\\\%mass=\boxed{\bold{11.33\%}}

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saveliy_v [14]

Hazardous materials are grouped into classes identifying their similarities in composition and structure.

<h3>Why hazardous materials are grouped into classes?</h3>

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8 0
1 year ago
Determine the poh of a 0.227 m c5h5n solution at 25°c. The kb of c5h5n is 1.7 × 10-9.
Juliette [100K]

4.71

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3 0
2 years ago
factors that affect the rate of a chemical reaction include which of the following? i. frequency of collisions of reactant parti
pickupchik [31]
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2 years ago
Please help with #2 and #3
Vaselesa [24]

Answer:

2. V_2=17L

3. V=82.9L

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

\frac{V_2}{T_2}=\frac{V_1}{T_1}

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

V_2=\frac{V_1T_2}{T_1}

Therefore, we plug in the given data to obtain:

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3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

V=\frac{nRT}{P}

Therefore, we plug in the data to obtain:

V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L

Best regards!

6 0
3 years ago
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alukav5142 [94]

Answer:

the moral amount of the gas.

5 0
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