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3 years ago

How do you prepare copper sulphate

1 answer:

Elena-2011 [213]3 years ago

4 0

Prepare a 1% copper sulfate solution. To make this solution, weigh 1 gram of copper sulfate (CuSO4 ·5H2O), dissolve in a small amount of distilled water in a 100 ml volumetric flask and bring to volume. Label this as 1% copper sulfate solution.

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Some one help me with this will give brainliest

Alona [7]

14. C
15. B
16. D
17. D
18. I am not sure sorry.<span />

3 0

3 years ago

Assume that instead of Taq polymerase, another polymerase, one that is very heat-sensitive, was used in a PCR. What differences

frez [133]

Answer:

Taq polymerase is a thermostable which states that it can even work at higher temperature. The main function of using this enzyme is that it is used to amplify the DNA which will help in producing ample amount of DNA sample.

The enzyme activity is temperature dependent. The denaturation and annealing steps of the PCR occurs at very high temperature.

Any other enzyme used at such an high temperature would have decreased the enzyme activity and the procedure would not be completed.

7 0

3 years ago

A 40g sample of polyisoprene, which has a specific heat capacity of f 1.88 J 1.°c, is dropped into an insulated container contai

Aneli [31]

Answer:

The equilibrium temperature of the water is 46.85 °C

Explanation:

Step 1: Data given

Mass of sample of polyisoprene = 40.0 grams

Specific heat capacity of polyisoprene = 1.88 J/g°C

Mass of water = 100 grams

Temperature of water = 55.0 °C

Pressure = 1 atm

The initial temperature of the polyisoprene is 1.5 °C

Step 2: Calculate the equilibrium temperature of the water

Heat lost = Heat gained

Qlost = - Qgained

Qwater = -Qpolyisoprene

Q = m*c*ΔT

m(water)*c(water)*ΔT(water) = -m(polyisoprene) * c(polyisoprene) * ΔT(polyisoprene)

⇒with m(water) = the mass of water = 100 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT = the change of temeprature = T2 - T1 = T2 - 55.0°C

⇒with m(polyisoprene) = the mass of polyisoprene = 40.0 grams

⇒with c(polyisoprene) = the specific heat of polyisoprene = 1.88 J/g°C

⇒with ΔT(polyisoprene) =the change of temperature = T2 -T1 = T2 - 1.5 °C

100 * 4.184 * (T2 - 55.0) = -40.0 * 1.88 * (T2 - 1.5)

418.4(T2 - 55.0) = -75.2(T2 - 1.5)

418.4 T2 - 23012 = -75.2T2+ 112.8

493.6 T2 = 23124.8

T2 = 46.85 °C

The equilibrium temperature of the water is 46.85 °C

6 0

3 years ago

What is the molarity of a solution that contains 25 g of HCl in 150 mL of solution? (The molar mass of HCl is 36.46 g/mol.)

lawyer [7]

Explanation:

Molarity is defined as number of moles of solute in liter of solution.

Whereas number of moles is equal to mass of given substance divided bu its molar mass.

Since molar mass of HCl is 36.46 g/mol. Therefore, calculate the number of moles of HCl as follows.

$Number of moles of HCl = \frac{mass of given}{Molar mass of HCl}$

= $\frac{25 g}{36.46 g/mol}$

= 0.686 mol

Hence, molarity of solution will be as follows.

Molarity = $\frac{no. of moles}{volume of solution}$

= $\frac{0.686 mol}{0.15 L}$

= 4.573 mol/L

Thus, we can conclude that molarity of a solution that contains 25 g of HCl in 150 mL of solution is 4.573 mol/L.

7 0

3 years ago

Read 2 more answers

A gas at constant temperature occupies a volume of 2.40 L and exerts a pressure of 0.85 atm. What volume will the gas occupy at

sesenic [268]

Answer:

13.6L

Explanation:

Given parameters:

Initial Volume V₁ = 2.4L

Initial pressure P₁ = 0.85atm

Final pressure P₂ = 0.15atm

Final volume V₂ = ?

Solution

To solve the problem we have to take the condition stated in the problem very carefully. It was stated that the gas was at a constant temperature.

According to Boyle's law" The volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant".

We can simply apply this law to solve the problem:

The law is mathematically expressed as P₁V₁ = P₂V₂

The unknown here is V₂ the final volume. We express it as the subject of the formula:

V₂ = $\frac{P_{1} V_{1} }{P_{2} }$

Now solving for the final volume, we have:

V₂ = $\frac{2.4 x 0.85}{0.15}$

V₂ = $\frac{2.04}{0.15}$ = 13.6L

4 0

4 years ago

How do you prepare copper sulphate​

How do you prepare copper sulphate