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IgorLugansk [536]
3 years ago
6

Why does one Oxygen atom bond with two Hydrogen atoms to form water? Why don't they bond in a different ratio?

Chemistry
1 answer:
Vladimir79 [104]3 years ago
4 0

This is because oxygen (2.8.6) requires two electrons on its valence shell to attain stable configuration (2.8.8). Hydrogen (1) on the other hand requires one electron on its valence shell to attain stable configuration (2). Therefore in a covalent bond, it requires two hydrogen and one oxygen to share electrons and achieve stable configuration.  

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A multistep reaction can only occur as fast as its slowest step. Therefore, it is the rate law of the slow step that determines
icang [17]

Answer:

b) rate = kAB.

Explanation:

Hello,

In this case, considering the given statement, we can notice that the rate law of the overall reaction will be determined for the slowest step, that is:

A+B \rightarrow AB\ \ (slow)

In such a way, we can infer that the rate law will contain both the concentration of A and B to the first power both, since their stoichiometric coefficients in the chemical equation are both one:

rate=k[A][B]

Thereby, answer is b) rate = kAB, that should be better rate = k[A][B] by expressing the concentrations.

Best regards.

8 0
3 years ago
Chloroform (CHCl3), an important solvent, is produced by a reaction between methane and chlorine. CH4(g) + 3 Cl2(g) CHCl3(g) + 3
PIT_PIT [208]
CH₄(g) + 3 Cl₂(g) → CHCl₃(g) + 3 HCl(g)
From the equation we notice that 1 mole of methane produces 1 mole of chloroform:
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?   g Methane → 37.5 g Chloroform
by cross multiplication:
= (16 * 37.5) / 119.38 = 5.0 g methane
4 0
3 years ago
A hot-air balloon is filled with air to a volume of 3000 m3 at 750 torr and 21°C. The air in the balloon is then heated to 60.°C
valina [46]

Answer: 1.47

Explanation:

The combined gas equation when pressure is constant:

\frac{V_1}{n_1T_1}=\frac{V_2}{n_2T_2}

where,

n_1 =original number of moles of air in the balloon = ?

n_2 =  number of moles of air in the heated balloon = ?

V_1 = initial volume of gas = 3000m^3

V_2 = final volume of gas = 5000m^3

T_1 = initial temperature of gas = 21^oC=273+21=294K

T_2 = final temperature of gas = 60^oC=273+60=333K

Now put all the given values in the above equation, we get the final pressure of gas.

\frac{3000}{n_1\times 294K}=\frac{5000}{n_2\times 333K}

\frac{n_2}{n_1}=1.47

Therefore, the  ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon is 1.47

5 0
3 years ago
Is this right?
telo118 [61]
Well I’m. Or to sure but it can’t be B because when you throw the ball the the kinetic energy is still increasing
3 0
3 years ago
Read 2 more answers
Please help me...
REY [17]

200 Fahrenheit 200 c  

7 0
4 years ago
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