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BlackZzzverrR [31]

3 years ago

10

chemist is using 387 mL of solution of acid in water if 13.3 of the solution is acid how many milliliters of acid are there roun

d answer to nearest tenth

1 answer:

Greeley [361]3 years ago

6 0

Answer:

V acid = 51.471 mL

Explanation:

%v/v = [(compoud volumen)/(solution volume)]×100

∴ V solution = 387 mL

∴ %v/v = 13.3 %

compund volume:

⇒ (V acid)/( V sln) = 0.133

⇒ V acid = (0.133)(387 mL)

⇒ V acid = 51.471 mL

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How will the volume of a gas be affected if the pressure is tripled, but the temperature remains the same?

dangina [55]

Answer:

Volume of the gass will decrease by three times of the original volume

Explanation:

Volume is inversly propotional to the pressure applied on it.

4 0

3 years ago

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Consider the reaction:

stich3 [128]

Answer:

$\large \boxed{\text{-851.4 kJ/mol}}$

Explanation:

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

$\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})$

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)

ΔfH°/kJ·mol⁻¹: 0 -824.3 -1675.7 0

$\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}$

7 0

3 years ago

In the formula for carbon tetrahydride, CH4, the percent composition of all elements is:

creativ13 [48]

C%=12/16*100=3/4*100=75%
H%=25%

8 0

2 years ago

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When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

3 0

2 years ago

A baby weighs 7 pounds, 4 ounces at birth and 7 pounds, I ounce at discharge. What percent of weight did the baby lose

Aloiza [94]

Answer:

2.6%

Explanation:

As, 1 ounce (oz) = 0.0625 pounds (lb)

Therefore, weight of baby at discharge = 7 lb,1 oz = 7+0.0625 lb = 7.0625 lb

Since, 1 oz = 0.0625 lb

⇒ 4 oz = 4×0.0625 = 0.25 lb

Therefore, weight of baby at birth = 7 lb,4 oz = 7+0.25 lb = 7.25 lb

The <u>amount of weight lost</u> is equal to the difference of weight of the baby at birth and discharge.

Therefore, <u>weight lost</u> = 7.25 lb - 7.0625 lb = <u>0.1875 lb</u>

Now, the <u>percentage of weight lost</u> by the baby is given by the amount of weight lost divided by the weight of the baby at birth.

Therefore, <u>the percentage of weight los</u>t = weight lost ÷ weight at birth = 0.1875 lb ÷ 7.25 lb × 100 = <u>2.6% </u>

6 0

3 years ago