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Fynjy0 [20]
3 years ago
15

If we add 20g of salt to 150g of water, what is the concentration by mass of the solution?

Chemistry
1 answer:
PIT_PIT [208]3 years ago
8 0

Answer:

11.76% by weight in salt

Explanation:

concentration = mass of solute / mass of solute + mass of solvent

mass of solute = 20 grams

mass of solvent = 150 grams

mass of solute + mass of solvent = 20g + 150g = 170g

concentration (weight %) =  (20g / 170g)100% = 11.76% by weight in salt.

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What is the amount of aluminum chloride produced from 40 moles of chlorine and excess aluminum?
Vera_Pavlovna [14]

Answer: Molar mass of Al = 26.98 g/mol

mass of Al = 12 g

mol of Al = (mass)/(molar mass)

= 12/26.98

= 0.4448 mol

According to balanced equation

mol of AlCl3 formed = moles of Al

= 0.4448 mol

Answer: 0.445 mol

hope this help boo❤️❤️❤️

Explanation:

3 0
3 years ago
The graph below shows the volume and the mass of four different substances at STP.
sladkih [1.3K]
The formula of density is mass / volume

This means that
- high mass, low volume = high density
- high mass, high volume = so-so
- low mass, high volume = low density

From the graph shown, 
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6 0
4 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:

V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L

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
Find the percent composition of each element in KMnO4
VARVARA [1.3K]
The way you want to find the percent composition would be by breaking down the problem like so:

K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999

Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO

Then you want to take each molar mass and then divide it 110.035 and multiply by 100

Ex. K = 39.098/ 110.035 and the multiply what you get by a 100

You do this for the other elements as well good luck!


6 0
3 years ago
Qué permite las reacciones de redox
liberstina [14]

Explanation:

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5 0
3 years ago
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