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Arada [10]
3 years ago
7

What is the percent by mass of the solution formed when 5.0 g of solute is dissolved in 40. g of water

Chemistry
1 answer:
d1i1m1o1n [39]3 years ago
7 0

Answer:

2

Explanation:

so let say 5.0/100×40

that is 2

thank you please follow me

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The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose 0.05 L of KOH w
strojnjashka [21]

From the chemical equation given:

H2SO4+2KOH--->K2SO4+2H2O

the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.

No. of moles of KOH = 2* no. of moles of H2SO4

=2*0.1*0.033

The concentration of KOH = no. of moles / volume

=2*0.1*0.033/0.05

=0.132M

3 0
3 years ago
Read 2 more answers
Interpret the following equation for a chemical reaction using the coefficients given:
Liono4ka [1.6K]

Answer:  1 molecule of H_2 reacts with 1 molecule of Cl_2 to give 2 molecules of HCl.

1 mole of H_2 reacts with 1 mole of Cl_2 to give 2 moles of HCl.

Explanation:

The given balanced reaction is:

H_2(g)+Cl_2(g)\rightarrow 2HCl(g)

On the particulate level :

1 molecule of H_2 reacts with 1 molecule of Cl_2 to give 2 molecules of HCl.

On molar level:

1 mole of H_2 reacts with 1 mole of Cl_2 to give 2 moles of HCl.

3 0
3 years ago
A 0.98 M solution of calcium chloride with a volume of 0.25 L is diluted to form a 0.74 M solution. What is the new volume of th
tankabanditka [31]

The new volume of the dilute solution is 0.33 L.

<u>Explanation:</u>

Using the law of Volumetric analysis, we can find the volume of the dilute solution from the stock solution or the concentrated solution of Calcium Chloride.

V1M1 = V2M2

V1 be the volume of the stock solution = 0.25 L

M1 being the molarity of the stock solution = 0.98 M

V2 be the volume of the dilute solution = ?

M2 being the molarity of the dilute solution = 0.74 M

We have to rearrange the above equation to find V2 as,

V2 = $\frac{V1M1}{M2}

Now plugin the values as,

V2 = $\frac{0.25 L \times 0.98 M }{0.74 M}

   = 0.33 L

So the new volume of the dilute solution is 0.33 L.

3 0
3 years ago
A mole of ethyl alcohol weighs 46 g. how many grams of ethyl alcohol are needed to produce 1 l of a 2-millimolar (2 mm) solution
Thepotemich [5.8K]
Answer: 92mg of ethyl alcohol 

2 milli molar of ethyl alcohol means 2 milli mole glucose for 1 liter of water. Since we want to make 1L of solution, then the amount of ethyl alcohol needed is: 2 milli mole/l x 1l= = 2 x 10^-3 mole

Ethyl alcohol molecular weight is 46g/mole, then 2 mili mole of glucose is= 2x 10^-3 mole x 46 grams/mole= 92 grams x 10^-3= 92mg
7 0
4 years ago
Determine the freezing point of a 0.765 m solution of nitrobenzene in naphthalene. (given: naphthalene Kf = 7.45o C Kg/ mole and
amid [387]

Answer: The freezing point of a 0.765 m solution of nitrobenzene in naphthalene is 74.6^0C

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(80.3-T_f)^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte such as nitrobenzene)

K_f = freezing point constant = 7.45^0C/m

m= molality  = 0.765

(80.3-T_f)^0C=1\times 7.45\times 0.765

T_f=74.6^0C

Thus the freezing point of a 0.765 m solution of nitrobenzene in naphthalene is 74.6^0C

7 0
3 years ago
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