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Andrew [12]
3 years ago
6

POV: You buy a composite cup at the store. Explain the reason for your decision.

Chemistry
2 answers:
Len [333]3 years ago
7 0

Answer:

i bought it cuz i need a cup?

Explanation:

dezoksy [38]3 years ago
7 0

Answer:

Best chemistry ever tho

Explanation:

LOL

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One symptom of carbon monoxide poisoning is the development of Mees lines on the fingernails and toenails.
satela [25.4K]
Mees' lines are white bands that form across your fingernails and are one of the symptoms of carbon monoxide poisoning.

Mees' lines are a sign that the growth of your nails temporarily stopped at one point. There are many causes to Mees' lines, and usually it indicates a toxicity. Carbon monoxide, along with other substances like arsenic, and selenium poisoning may cause them. 


3 0
3 years ago
Read 2 more answers
The atomic mass of chlorine is listed as 35.454 amu on the periodic table. It has two naturally occurring isotopes, Cl-35 and Cl
Fynjy0 [20]
X=percentage of Cl-35
(1-x)=percentage of Cl-35

35(x)+37(1-x)=35.454
35x+37-37x=35.454
-2x=35.454-37
-2x=-1.546
x=-1.546/(-2)
x=0.773

0.773=77.3/100=77.3%

(1-x)=1-0.773=0.227

0.227=22.7/100=22.7%

answer: the percentages are: 77.3% of Cl-35 and 22.7% of Cl-37. 

7 0
3 years ago
a sample of the hydrate of sodium carbonate has a mass of 8.85 g. it loses 1.28 g when heated. find the molar ratio of this comp
xenn [34]

Answer:

Molar ratio of the compound is 1:1 and the type of hydrate is Mono hydrate.

Explanation:

From the given,

Mass of sodium carbonate Na_{2}CO_{3}.XH_{2}O = 8.85 g

Loss mass H_{2}O = 1.28 g

Actual weight of sodium carbonate = 8.85 g - 1.28 g = 7.57 g

7.57 g Na_{2}CO_{3} \times \frac{1mol}{106 g} =\frac{0.0714}{0.0714}=1

1.28g H_{2}O \times \frac{1mol}{18 g} =\frac{0.0711}{0.0714}=1

Therefore, the compound has only one water molecule.

Molecular formula of the compound is Na_{2}CO_{3}.H_{2}O an name of the compound is <u>sodium carbonate mono hydrate.</u>

Hence, the type of the compound is Mono hydrate.

8 0
3 years ago
If 20.0g of CO2 and 4.4g of CO2
Ksivusya [100]

The given question is incorrect. The correct question is as follows.

If 20.0 g of O_{2} and 4.4 g of CO_{2}  are placed in a 5.00 L container at 21^{o}C, what is  the pressure of this mixture of gases?

Explanation:

As we know that number of moles equal to the mass of substance divided by its molar mass.

Mathematically,   No. of moles = \frac{\text{mass}}{\text{molar mass}}

Hence, we will calculate the moles of oxygen as follows.

       No. of moles = \frac{\text{mass}}{\text{molar mass}}

     Moles of O_{2} = \frac{20.0 g}{32 g/mol}

                            = 0.625 moles

Now,   moles of CO_{2} = \frac{4.4 g}{44 g/mol}

                                      = 0.1 moles

Therefore, total number of moles present are as follows.

Total moles = moles of O_{2} + moles of CO_{2}

                    = 0.625 + 0.1

                    = 0.725 moles

And, total temperature  will be:

                    T = (21 + 273) K = 294 K

According to ideal gas equation,  

                         PV = nRT

Now, putting the given values into the above formula as follows.

                P = \frac{nRT}{V}

                   = \frac{0.725 mol \times 0.08206 Latm/mol K \times 294 K}{5.00 L}

                    = \frac{17.491089}{5} atm

                    = 3.498 atm

or,                = 3.50 atm (approx)

Therefore, we can conclude that the pressure of this mixture of gases is 3.50 atm.

4 0
3 years ago
Draw the structure of a compound with the molecular formula CgH1002 that exhibits the following spectral data.
Gnesinka [82]

Answer:

The answer you are looking for is A

7 0
3 years ago
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