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Ghella [55]
3 years ago
7

The balanced chemical equation for the reaction of copper (Cu) and silver nitrate (AgNO3) is shown below. Cu + 2AgNO3 mc001-1.jp

g 2Ag + Cu(NO3)2 How many moles of copper must react to form 3.50 mol of Ag?
Chemistry
2 answers:
sergij07 [2.7K]3 years ago
8 0
The balanced reaction is:

<span>Cu + 2AgNO3 = 2Ag + Cu(NO3)2

We are givent the amount of silver to be produced. This will be the starting point. Calculations are as follows:

</span><span> 3.50 mol Ag ( 1 mol Cu / 2 mol Ag ) = 1.75 mol Cu
</span>
Therefore, we need 1.75 mol of copper in order to produce 3.50 mol of silver.
nirvana33 [79]3 years ago
5 0

Answer:

1.75 moles of copper must react to form 3.50 mol of Ag.

Explanation:

Cu + 2AgNO_3\rightarrow 2Ag + Cu(NO_3)_2

According to reaction 2 moles of silver are obtained from 1 mole of copper.

Then 3.50 mole of Silver will be obtained from:

\frac{1}{2}\times 3.5 mol=1.75 mol

1.75 moles of copper must react to form 3.50 mol of Ag.

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You need 455 ml of a 75% alcohol solution. on hand, you have a 25% alcohol mixture. you also have a 90% alcohol mixture. how muc
marishachu [46]
X ml - <span>25% alcohol mixture
y ml - </span><span>90% alcohol mixture

x+y = 455 

0.25x ml alcohol in </span>x ml of  25% alcohol mixture
0.9y ml alcohol in y ml of  90% alcohol mixture
0.75*455= 341.25 ml alcohol in 455 ml of  75% alcohol mixture

0.25x+0.9y=341.25 

System of equations:

x+y = 455 /*(-0.25) ------> -0.25x-0.25y = -0.25*455 
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-0.25x-0.25y=-113.75
0.25x+0.9y=341.25    Add both equations

0.25x+0.9y-0.25x-0.25y=341.25-113.75
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8 0
3 years ago
The data below shows the change in concentration of dinitrogen pentoxide over time, at 330 K, according to the following process
tensa zangetsu [6.8K]

Answer: a) 1.7\times 10^{-4}

b) 3.4\times 10^{-4}

Explanation:

The reaction is :

2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)

Rate = Rate of disappearance of N_2O_5 = Rate of appearance of NO_2

Rate =  -\frac{d[N_2O_5]}{2dt} = \frac{d[NO_2]}{4dt}

Rate of disappearance of N_2O_5 = \frac{\text {change in concentration}}{time} = \frac{0.100-0.066}{200-0}=1.7\times 10^{-4}

a) Rate of disappearance of N_2O_5 = -\frac{d[N_2O_5]}{2dt}

Rate of appearance of NO_2 = \frac{d[NO_2]}{4dt}

b) Rate of appearance of NO_2 =  \frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}

8 0
3 years ago
Which branch of chemistry specifically deals with quantitative analysis?
LUCKY_DIMON [66]

Answer:

analyitical

Explanation:

4 0
4 years ago
Read 2 more answers
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