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2 years ago

So when dealing with molar mass when do you know to round up

1 answer:

8 0

The first rule is to use four digits in atomic weights and consider the calculated formula or molar mass as given with four significant digits. This rule is appropriate for beginners but has some disadvantages.

The second rule is to use all digits in the recommended atomic weights and to estimate the accuracy of the result from the number of decimal places or significant digits in these.

The third rule is to use all digits in the recommended atomic weights and to calculate the absolute maximum uncertainty of the result from the recommended uncertainties of the atomic weights. The uncertainty is rounded up to one (two) significant digit(s), and the formula or molar mass is rounded to the same absolute accuracy as the rounded uncertainty.

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A hypothetical main group element E reacts with chlorine to form an ionic compound with the formula ECl. The element is a member

Ronch [10]

It would be 1A bc then the +1 charge will cancel out chlorine’s -1 charge

7 0

3 years ago

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

3 years ago

1. Which is not true of binary compounds?

nignag [31]

Pretty sure it's b but not an definitely

4 0

3 years ago

Read 2 more answers

Water glass is found in

klemol [59]

Answer:

liquid form

Explanation:

am i right? if right like

5 0

3 years ago

Read 2 more answers

4 NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Mariulka [41]

Answer:

0.9 moles of water

Explanation:

Use mole ratios:

5 : 6

divide by 5 on both sides

1 : 1.2

multiply by 0.75 on both sides

0.75 : 0.9

So the result is 0.9 moles of water

(Please correct me if I'm wrong)

5 0

2 years ago