All categories

2 years ago

How many molecules are there in 322 grams of Cu(NO3)2

Chemistry

1 answer:

mr_godi [17]2 years ago

3 0

<span>cu=1*63g=63g
n=2*14g=28g
o=6*16g=96g
total=187g/mol
122(1 mol/187)(6.02*10^23)= 3.92*10^23 </span><span>
</span>

You might be interested in

What is the mole to mole ratio of LiOH to HBr for the following reaction?

umka21 [38]

Explanation:

so moles = mass ÷ mr (1+ 79.9)

so 10.00g ÷ 80.9

which is 0.1236093943

so to 3 S.F is 0.124 moles

also there is 1 to 1 ratio for LiOH to HBr

hope this helps :)

8 0

2 years ago

Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac

GalinKa [24]

<h2>a) The rate at which $NO_2$ is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen $O_2$ is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of $NO$ = $-\frac{1d[NO]}{2dt}$ = 0.066 M/s

Rate in terms of disappearance of $O_2$ = $-\frac{1d[O_2]}{dt}$

Rate in terms of appearance of $NO_2$ = $\frac{1d[NO_2]}{2dt}$

1. The rate of formation of $NO_2$

$-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}$

$\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s$

2. The rate of disappearance of $O_2$

$-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}$

$-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s$

Learn more about rate law

brainly.com/question/13019661

https://brainly.in/question/1297322

7 0

2 years ago

A study of the decomposition reaction 3RS2  3R + 6S yields the following initial rate dat

shutvik [7]

Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.

3 0

2 years ago

Calculate the percentage by mass of the indicated element in the following compounds. Hydrogen in ascorbic acid, HC6H7O6, also k

Scorpion4ik [409]

Answer: The mass percent of hydrogen in ascorbic acid is 4.5 %

Explanation:

In $C_6H_8O_6$ , there are 6 carbon atoms, 8 hydrogen atoms and 6 oxygen atoms.