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andrew11 [14]
3 years ago
15

Bauxite is an ore that contains the element aluminum. If you obtained 108 grams of aluminum from an ore sample that initially we

ighed 204 grams, what is the mass percent of aluminum in this bauxite ore
Chemistry
1 answer:
Stels [109]3 years ago
8 0

Answer:

53% aluminium is present.

Explanation:

Given data:

Mass of aluminium in bauxite = 108 g

Total mass of sample = 204 g

Percentage of aluminium = ?

Solution:

Formula:

Percentage = Mass of aluminium / total mass of sample× 100

Now we will put the values in formula:

Percentage = 108 g/204 × 100

Percentage = 0.53  × 100

Percentage = 53%

So in given 204 g of bauxite there is 53% aluminium is present.

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At 47c a gas has a pressure of 140kpa. The gas is cooled until the pressure decreases to 105kpa. If the volume remains constant,
IgorC [24]

Answer:

The final temperature is equal to 240 K or -33.15°C

Explanation:

Given that,

Initial temperature of the gas, T₁ = 47°C = 320 K

Initial pressure, P₁ = 140 kpa

Final pressure, P₂ = 105 kpa

We need to find the final temperature if the volume remains constant.  The relation between temperature and pressure is given by :

P\propto T

or

\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{P_2T_1}{P_1}\\\\T_2=\dfrac{105\times 320}{140}\\\\T_2=240\ K\\\\T_2=-33.15^{\circ} C

So, the final temperature is equal to 240 K or -33.15°C.

7 0
3 years ago
a gas has a volume of 2.00 at 323k and 3.00 ATM What would be the new volume if the temperature is changed to 273 K in the press
Ber [7]

Answer:

V_2 = 5.07L

Explanation:

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\\V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{3.00atm \cdot 2.00L \cdot 273K}{323K \cdot 1.00atm} = 5.07L

6 0
2 years ago
Convert 732.0 mmHg to atm
enot [183]

Answer:

The answer is

<h2>0.95 atm</h2>

Explanation:

To solve the question we use the following conversion

That's

1 mmHg \cong 0.0013 atm

So we have

If 1 mmHg \cong 0.0013 atm

Then 732 mmHg will be

732 × 0.0013 atm

We have the final answer as

<h3>0.95 atm</h3>

Hope this helps you

5 0
4 years ago
Calculate the pH of a buffer solution prepared by mixing 60.0 mL of 1.00 M lactic acid and 25.0 mL of 1.00 M sodium lactate.
marshall27 [118]
This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:

pH = pKa + log[(salt/acid]

Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid). 

Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L

Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M

Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48
4 0
3 years ago
Is boiling eggs chemical or physical practice?
Elden [556K]

Answer:

chemical

Explanation:

because heat is being taken to the egg

3 0
3 years ago
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