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kondor19780726 [428]
3 years ago
13

What is the solubility of 2 kilograms of potassium chloride (KCl) dissolved into 750 cm3 of water? A. 0.375g/dm3 B. 1500g/dm3 C.

2.67g/dm3 D. 2666.67g/dm3
Chemistry
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

D. 2666.67g/dm3

Explanation:

Hello,

In this case, we compute the solubility as:

solubility=\frac{m}{V}

Thus, we compute:

solubility=\frac{2kg}{750cm^3}*\frac{1000g}{1kg} *\frac{1000cm^3}{1dm^3}\\  \\solubility=2666.7\frac{g}{dm^3}

Therefore, answer is D. 2666.67g/dm3.

Best regards.

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Calculate how many atoms are in 98.6 g Carbon (C)?
Goryan [66]
Number of atoms are calculated using a number known as "Avogadro's number."
Okay, lets see a breakdown of this. Let's say that you are given an amount of grams of a substance. For this case, lets say that that substance is Carbon (C). And, lets assume that you are given 4.01 g of Carbon, and you are tasked to find the number of atoms in that mass of Carbon. The breakdown would be as follows, with dimensional anaysis:
4.01
g Carbon
(
1
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12.01
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(
6.022
⋅
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23
a
t
m
s
C
a
r
b
o
n
1
mol Carbon
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=
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10
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atms Carbon
Basically, I first wrote down the amount in grams, and I used the molar mass of Carbon (which can be found on the periodic table under Carbon) 12.01 g/mol to convert 4.01 g of Carbon to moles of Carbon. Then, I used "Avogadro's Number", or
6.022
⋅
10
23
atoms per mole
to convert the mole amount to atoms of Carbon.
The process should be very similar with other such atoms, just make sure to keep your periodic table and calculator handy.
I hope that helps!
5 0
4 years ago
What is the stock and classical name for CrBr3?
madam [21]

Answer:

The stock name (CrBr3) :  <em>chromium(III) sulfide</em>

The classical name (CrBr3) : <em>chromic bromide</em>

3 0
4 years ago
An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 15.71 g CaCl2. After performing the experim
OlgaM077 [116]

Answer:

93.15 %

Explanation:

We have to start with the chemical reaction:

CaCl_2~+~Na_2CO_3~->~CaCO_3~+~NaCl

Now, we can balance the reaction:

CaCl_2~+~Na_2CO_3~->~CaCO_3~+~2NaCl

Our initial data are the 15.71 g of CaCl_2, so we have to do the following steps:

1) <u>Convert from grams to moles of CaCl_2 using the molar mass (110.98 g/mol).</u>

2) <u>Convert from moles of CaCl_2 to moles of CaCO_3 using the molar ratio. ( 1 mol CaCl_2= 1 mol of CaCO_3).</u>

3) <u>Convert from moles of CaCO_3 to grams of CaCO_3 using the molar mass. (100 g/mol).</u>

15.71~g~CaCl_2\frac{1~mol~CaCl_2}{110.98~g~CaCl_2}\frac{1~mol~CaCO_3}{1~mol~CaCl_2}\frac{100~g~CaCO_3}{1~mol~CaCO_3}=14.16~g~CaCO_3

Finally, we can calculate the yield percent:

%~=~\frac{13.19~g~CaCO_3}{14.16~g~CaCO_3}*100=93.15~%

I hope it helps!

5 0
4 years ago
What kind of charge would the oxygen atom have compared to the hydrogen atoms
Oliga [24]
It should be charge 2-they lose electrons
8 0
4 years ago
Which statement about polymers is true? A. Polymers are substances that have relatively large molecules. B. Oil and gas are the
igor_vitrenko [27]

Answer:

A.) Polymers are substances that have relatively large molecules

3 0
4 years ago
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