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Scorpion4ik [409]
1 year ago
11

A chemist weighed out 5.14 g of a mixture containing unknown amounts of bao (s) and cao (s) and placed the sample in a 1.50-l fl

ask containing co2 (g) at 30.0 oc and 750. torr. after the reaction to form baco3 (s) and caco3 (s) was completed, the pressure of co2 (g) remaining was 230. torr. calculate the mass percentages of cao (s) and bao (s) in the mixture. 86.6 % bao; 13.4 % cao
Chemistry
1 answer:
lora16 [44]1 year ago
7 0

Let the mass of CaO = x grams

So mass of BaO = 5.14 -x grams

moles of CaO = mass / molar mass = x / 56

Moles of BaO =  mass /  molar mass = 5.14-x / 153

initial moles of CO2 = PV / RT = 750 X 1.50 / 760 X 0.0821 X 303 = 0.06

final mole sof CO2 = PV / RT = 230 X 1.50 / 760 X 0.0821 X 303 = 0.018

So moles of BaCO3 and CaCO3 formed = 0.06 - 0.018 = 0.042 moles

x / 56 + (5.14-x) /153 = 0.042

on solving

x = 0.68

So mass of CaO = 0.68 g

So percentage of CaO = 0.68 X 100 / 5.14 = 13.4 %

Percentage of BaO = 86.6%


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This problem has been solved! See the answer Determine Um (mode ), average U, and Urms for a group of ten automobiles clocked by
lara31 [8.8K]

Explanation:

The mode is the most common number.

Um = 55

The mean is the sum of the numbers divided by the quantity.

Uavg = (38 + 44 + 45 + 48 + 50 + 55 + 55 + 57 + 58 + 60) / 10

Uavg = 51

The RMS (root mean square) is the square root of the sum of the squares of the numbers divided by the quantity.

Urms = √[(38² + 44² + 45² + 48² + 50² + 55² + 55² + 57² + 58² + 60²) / 10]

Urms = 51.451

4 0
2 years ago
A 0.300 g piece of copper is heated and fashioned into a bracelet. The amount of energy transferred by heat to the copper is 66,
Alja [10]

Answer:

X=600 degrees celsius

Explanation:

We can evaluate by using the equation ==> Q=m(c)(delta T)

-We know the mass is 0.3 g

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-C= 390 j/g 0C

-Therefore, we need to solve for delta T (which will be represented by x)

 Q= mct

66,300 J= 0.3 g (390 J/g 0C) (t)

The unit of measurement (g-grams) can be crossed out on the left side of our expression.

66,300 J=0.3(390 J 0C) (t)

-If we convert into a proportional relationship:

66,300 J                117 J/0C x

------------       =        -----------

117 J/0C                 117 J/0C

-Joules on both sides can be crossed off

*117 J/0C came from the product in terms of 0.3(390 J 0C)

*If we find the quotient of 66,300 J and 117 J/0C, we get 566.6 repeating

-You may keep the answer as it is, although for simplicity reasons, we can round to the nearest hundred.

-Therefore, x=600.

-----Remember, x represented the delta t in our equation. (I made notice of that when evaluating).

Hope this helps! :)

3 0
1 year ago
THESE QUESTIONS ARE ELECTRICIAN REALATED
Vladimir [108]

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It is a technique typically engaged in electrical power and electronic devices, wherein the devices are run at less than their rated maximum power degeneracy, taking into account the case or body temperature, the ambient temperature and the type of cooling mechanism used.

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<span>16.  </span>D.

<span>17.  </span>C.

Table 310-16 NEC

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4 0
2 years ago
Read 2 more answers
A hot gas flowing through a pipeline can be considered as a:________
k0ka [10]

Answer:

B) irreversible process

Explanation:

The process given here is irreversible.

7 0
1 year ago
Which of the following is not the same as 0.032 liters?
sineoko [7]
This question needs the answer choices.

I found these choices for you:
<span>0.00032hL
320cL
32mL

Then  you need to make the conversion of 0.032 liters to hectoliters, centiliters and milimilters to check which is not equivalent.

1) 0.032 liters to hectoliters:

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2) 0.032 liter to centiliters:

    0.032 liter * 100 centiliters / 1 liter = 3.2 centiliter

3) 0.032 liter to mililiter:

    0.032 liter * 1000 mililiter / liter = 32 mililiter

Then, the answer is 320 cL: 320 cL is not the same as 0.032 liters

</span>
3 0
2 years ago
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