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strojnjashka [21]
4 years ago
13

To prepare for a laboratory?

Chemistry
1 answer:
yKpoI14uk [10]4 years ago
6 0

Answer:

You have to prepare for the lab (Materials, work, paper etc.)

Set up the lab know where the lab will be taking place

Read thru the experiment before doing the lab

Make a hypothesis

Write down notes, observations, measures anything important to help with the lab!

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Calculate the number of moles in 79.17 g of magnesium chloride
natulia [17]
There would be 0.8315 moles
7 0
3 years ago
If 1.85 g Al reacts with an excess copper(II) sulphate and the percentage yield of Cu is 56.6%, what mass of Cu is produced
timofeeve [1]
We have a balanced equation: 
2 Al+ 3 CuSO4 ⇒ Al2(SO4)3+ 3 Cu

1.85 g Al* (1 mol Al/ 26.98 g Al)* (3 mol Cu/ 2 mol Al)* ( 63.55 g Cu/ 1 mol Cu)= 6.54 g Cu.
(Note that the units cancel out so you get the answer)

6.54 g Cu* (56.6/100)= 3.70 g Cu.

The final answer is 3.70 g Cu~
7 0
4 years ago
Read 2 more answers
The mole fraction of he in a gaseous solution prepared from 4.0 g of he, 6.5 g of ar, and 10.0 g of ne is __________.
solniwko [45]

Given that there is 4.0 g of He, 6.5 g of Ar, and 10.0 g of Ne :

So, first all these masses are converted into moles:

Numer of moles= \frac{Given mass}{Molar mass}


4.0 g of He =\frac{4 g}{4 \frac{g}{mol}}=1 mol


6.5 g of Ar =\frac{6.5 g}{39.95 \frac{g}{mol}} = 0.1627 mol


10.0 g of Ne =\frac{10 g}{20.18 \frac{g}{mol}} = 0.4955 mol


Total Number of moles  = 1 + 0.1627 + 0.4955 = 1.658 mol


Mole fraction of He in substance =\frac{Number of moles of He }{Total number of moles}

                                                = \frac{1}{1.695}=0.6

3 0
3 years ago
The volume of a cylinder is given by the formula?
ipn [44]

Answer:

V=πr²h

Explanation:

3 0
3 years ago
N2O5(g)→NO3(g)+NO2(g) Time (s) [N2O5] (M) 0 1.000 25 0.822 50 0.677 75 0.557 100 0.458 125 0.377 150 0.310 175 0.255 200 0.210 D
sashaice [31]

Answer:

The order of the reaction is 1.

Explanation:

N_2O_5(g)\rightarrow NO_3(g)+NO_2(g)

The rate law of the reaction ;

R=k[N_2O_5]^x

The rate of the reaction from T = 0 s to 25 s, [N_2O_5]=1.000 M to [N_2O_5]=0.822 M respectively.

R=-\frac{0.822 M-1.000 M}{25s-0 s}=0.00712 M/s

0.00712 M/s=k[0.822 M]^x ..[1]

The rate of the reaction from T = 25 s to 50 s, [N_2O_5]=0.822 M to [N_2O_5]=0.677 M respectively.

R=-\frac{0.677 M-0.822 M}{50s-25 s}=0.0058 M/s

0.00646 M/s=k[0.677 M]^x ..[2]

[1] ÷ [2]

\frac{0.00712 M/s}{0.0058 M/s}=\frac{k[0.822 M]^x}{k[0.677 M]^x}

Solving for x:

x = 1.05 ≈ 1

The rate law of the reaction ;

R=k[N_2O_5]^1

The order of the reaction is 1.

5 0
4 years ago
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