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anygoal [31]
4 years ago
6

If 29.4 mL of ethanol is dissolved in water to make 359 mL of solution, what is the concentration expressed in volume/volume % o

f the solute?
Chemistry
1 answer:
Sveta_85 [38]4 years ago
6 0

Answer: 8.2\%

Explanation:- Volume percentage is the ratio of volume of solute to the volume of solution defined in terms of percentage.

{\text {volume percentage}}=\frac{\text {volume of solute}}{\text {volume of solution}}\times 100\%

Given: volume of solute = 29.4 ml

Volume of solution= 359 ml

{\text {volume percentage}=\frac{29.4}{359}\times 100\%=8.2\%


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At 25.0∘C, the molar solubility of barium chromate in water is 1.10×10−5 M . Calculate the solubility in grams per liter.
Lunna [17]
Hey there!

Molar mass :
<span>
 ‎BaCrO</span>4 =  ‎253.319 g/mol

(g/L) = molar solubility * molar mass

(g/L) =  ( 1.10 x 10⁻⁵ ) * 253.319

=> 2.79x10⁻³ g/L
7 0
3 years ago
What does conserving mass mean in a chemical equation?
AleksandrR [38]
I would say that what conserving mass in a chemical equation means is that C. There is equal number of each type of atom on the reactant side and product side.
The equation is equal and no material is lost or gained.
4 0
3 years ago
Read 2 more answers
The element silver (Ag) has two naturally occurring isotopes: 109Ag and 107Ag with a mass of 106.905 amu. Silver consists of 51.
algol [13]

Answer:

108.76 amu is the mass of 109Ag

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 109Ag:

% = 51.82 %

Mass = x amu

For second isotope, 107Ag:

% = 100  - x   = 100 - 51.82 % = 48.18 %

Mass = 106.905 amu

Given, Average Mass = 107.868 amu

Thus,  

107.868=\frac{51.82}{100}\times {x}+\frac{48.18}{100}\times {106.905}

Solving for x, we get that:

x = 108.76 amu

<u>108.76 amu is the mass of 109Ag</u>

6 0
3 years ago
Calculate the volume in liters of a 29.8 g/dL nickel(II) chloride solution that contains 131. G of nickel(II) chloride . Be sure
ohaa [14]

Answer:

0.44 L.

Explanation:

Density of nickel(II) chloride = 29.8 g/dL.

Mass of nickel(II) chloride = 131 g

Volume of nickel(II) chloride =?

Next, we shall convert 29.8 g/dL to g/L. This can be obtained as follow:

Recall:

1 g/dL = 10 g/L

Therefore,

29.8 g/dL = 29.8 x 10 = 298 g/L

Therefore, 29.8 g/dL is equivalent to 298 g/L.

Finally, we shall determine the volume of nickel(II) chloride as follow:

Density of nickel(II) chloride = 298 g/L

Mass of nickel(II) chloride = 131 g

Volume of nickel(II) chloride =?

Density = mass /volume

298 = 131/Volume

Cross multiply

298 x Volume = 131

Divide both side by 298

Volume = 131/298

Volume = 0.44 L

Therefore, the volume of of nickel(II) chloride is 0.44 L

3 0
3 years ago
In a titration, the point at which one drop of base turns the acid indicator a pink color that lasts for 30 seconds is called th
xxTIMURxx [149]
The point at which one drop of base turns the acid indicator into a pink color that lasts for thirty seconds in doing titration is called the end point or the equivalence point.

End point or the equivalence point is the one responsible for the pink color that lasts for thirty seconds.
8 0
3 years ago
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