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BARSIC [14]
3 years ago
12

Methyl bromide (CH3Br) a. is extremely flammable b. is of greater concern from its oral toxicity than from its inhalation toxici

ty c. is a liquid used primarily as a fumigant d. has essentially no warning properties, even at physiologically hazardous concentrations
Chemistry
1 answer:
Mekhanik [1.2K]3 years ago
4 0

Answer:

I believe it is C.

Explanation:

Hope my answer has helped you and if not i'm sorry.

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Please hurry!!!!!
maksim [4K]
Light. ever heard of moon cycles? so d and c are out of the question. and although almost non-existent, the moon does have gravity. only thing the moon doesn't have is light emitting from itself
6 0
3 years ago
What is the mass of sulfuric acid, H2SO4, that is contained in 2.0 L of a 5.85 M solution ?
Elza [17]
D.-98.07 Grams this means that every mole of sulfuric acid has a mass of 98.0795g.
4 0
3 years ago
Read 2 more answers
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
suppose the burner under the pan of boiling water is turned to a higher settings. how will this affect the temperature of the wa
andrezito [222]

Answer:

Boiling hot?

Explanation:

5 0
3 years ago
Someone please help me
ehidna [41]
I hop this will help u

8 0
3 years ago
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