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Stells [14]
3 years ago
9

Im the only metal who is a liquid at room temperature

Chemistry
2 answers:
mylen [45]3 years ago
5 0
The only metal that is a liquid at room temperature is mercury!!
belka [17]3 years ago
3 0
The answer is mercury
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Complete combustion of 8.10 g of a hydrocarbon produced 25.9 g of CO2 and 9.27 g of H2O. What is the empirical formula for the h
balu736 [363]

CxHy     +  O2    -->    x CO2     +    y/2  H2O

 

Find the moles of CO2 :     18.9g  /  44 g/mol   =    .430 mol CO2   = .430 mol of C in compound

Find the moles of H2O:      5.79g / 18 g/mol     =     .322 mol H2O   = .166 mol of H in compound

 

Find the mass of C and H in the compound:

                             .430mol  x 12  =  5.16 g C

                              .166mol  x 1g   = .166g H   

 

When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.

Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).

In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).

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4 years ago
Which is not on a station model? *
Fiesta28 [93]
The answer is fronts
6 0
3 years ago
A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
Chemical reactions are _____.
lana66690 [7]

Answer:

I think the answer is B)the processes in which molecules interact to form new subtances.

7 0
3 years ago
Which of the following hypotheses is testable by using the scientific method?
Paraphin [41]

Answer:

D

Explanation:

Because it is the only was that you can test using the scientific method.

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