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Debora [2.8K]
2 years ago
10

explain observation made when anhydrous calcium chloride and anhydrous copper (ii) sulphate are separately exposed to the atmosp

here for about two days​
Chemistry
1 answer:
Burka [1]2 years ago
6 0

Answer:

Anhydrous calcium chloride  dissolves and becomes liquid

Anhydrous copper (ii) sulphate will produce crystal particles

Explanation:

Anhydrous calcium chloride is deliquescent and hence when it is exposed to air, it absorbs water from air. After absorbing water, it dissolves and after some time a pool of clear liquid appears.

Anhydrous copper (ii) sulphate will form crystal structures  and the following reaction will takes place

CuSO4 + 5 H20 --> CuSO4.5H2O

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The average American is looking at their phone for 6.25 hours a day. How many hours will the average American spend on their pho
julia-pushkina [17]

Answer:

177253.125hrs

Explanation:

Given parameters:

  Number of hours spend per day on phone  = 6.25hrs

 Unknown:

Average hours pend by an average American if they lived for 77.7yrs

Solution:

Let us find the number of hours an average American spends on their phone per year;

    there are 365days in a year

 Number of hours  = 365 x 6.25  = 2,281.25hrs

Now, if they lived for 77.7yrs;

  They will spend  = 77.7 x 2,281.25 = 177253.125hrs

8 0
2 years ago
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Vikki [24]
Yes what is this tho like
8 0
2 years ago
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If 36 000 kg of full cream milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream
Yuri [45]

Answer:

B=5522.33kg/h

C=478.11kg/h

Explanation:

Hi! It's a mass balance. First we have to determine the inflow.

mass flow rate = 36000kg / 6h = 6000kg / h

We define the input variable

- input flow (A) = 6000kg / h

-XgA = percentage of fat in A = 0.04

We define output variables.

- skim milk (B)

-creme (C)

-XgB = fat percentage at B = 0.0045

-XgC = percentage of fat in C = 0.45

Then we can start with the balance.

As a general rule, the mass balance is:

Input = Output

Balance sheet

1) A = B + C

Fat balance

2) A * XgA = B * XgB + C * XgC

Now we can solve.

We replace and clear B in equation 2

6000kg / h * 0.04 = B * 0.0045 + C * 0.45

B = (240kg / h) /0.045-C*0.45/0.0045

3) B = 53333.33kg / h-C * 100

We replace equation 3 in 1 and clear C

A = B + C

6000kg / h = 53333.33kg / h-C * 100 + C

C=(6000kg/h-53333.33kg/h)/(-99)

C=478.11kg/h

We replace C in equation 3 and calculate B

B = 53333.33kg / h-478.11kg/h * 100

B=5522.33kg/h

Then we have the values ​​of the outflows.

C=478.11kg/h

B=5522.33kg/h

7 0
3 years ago
I need help with stoichometrey
alukav5142 [94]
I can help you! What is your question?
8 0
3 years ago
The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.
harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})

= 70molCO_2

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})

= 87.5molO_2

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

84.0molO_2(\frac{4molO_2}{5molO_2})

= 67.2molCO_2

7 0
3 years ago
Read 2 more answers
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