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astraxan [27]
3 years ago
9

CAN SOMEBODY PLEASE HELP ME ON CHEMISTRY OF 8 th Grade on my exam

Chemistry
1 answer:
il63 [147K]3 years ago
7 0
No screenshots of anything you need ?
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What is a solution?
Keith_Richards [23]

Answer:

A !!

it a mix where everything is combined together

5 0
2 years ago
Read 2 more answers
Estimate the molar mass of a gas that effuses at 1.80 times the effusion rate of carbon dioxide. answer in units of g/mol.
Effectus [21]
From Grahams Law the rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure.
Therefore; R1/R2 = √mm2/√mm1
The molecular mass of Carbon dioxide is 44 g
Hence;  1.8 = √(44/x
             3.24 = 44/x
                x = 44/3.24
                   = 13.58 
Therefore, the molar mass of the other gas is 13.58 g/mol
           
3 0
3 years ago
1. Why don't scientists stop with the first step in the Scientific Method, making observations? A. They don't trust their own ob
natulia [17]

Answer:

Explanation:

Scientists don't stop with the first step of their experiment because they want other scientists' opinions because they may not trust their own observations.

                     OR

Scientists don't stop with the first step of their experiment because they would rather plan and run experiments than just observe the world around them    

       

Hope any one of these helps you

5 0
3 years ago
The blank which are represented by a single uppercase letter,or represented by an uppercase letter followed by a lowercase lette
melomori [17]

Answer: I believe it’s element symbols

Explanation: I hope we both get it right

7 0
3 years ago
1. Mixing Water at Two Temperatures
aksik [14]

Answer:

T_f=72.5\°C

Explanation:

Hello!

In this case, since this a problem in which the cold water is heated by the hot water, we can write:

Q_{hot}+Q_{cold}=0

Thus, by plugging in the mass, specific heat and temperatures, we obtain:

m_{hot}C_{hot}(T_f-T_{hot})+m_{cold}C_{cold}(T_f-T_{cold})=0

Now, we can also write:

m_{hot}(T_f-T_{hot})+m_{cold}(T_f-T_{cold})=0

Then, after applying some algebra, it is possible to obtain:

T_f=\frac{m_{hot}T_{hot}+m_{cold}T_{cold}}{m_{hot}+m_{cold}}

If we plug in, we obtain:

T_f=\frac{350.0g*95.0\°C+150.0g*20.0\°C}{350.0g+150.0g}

T_f=72.5\°C

Best regards!

8 0
3 years ago
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