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stiv31 [10]
3 years ago
6

A solar heating specialist is considering paraffin (Cp = 2.90 J/g O C) as a storage material for heat. How many kg of paraffin w

ould be needed to collect the amount of energy as 4.73 x 10 3 kg of water?
Chemistry
1 answer:
Stels [109]3 years ago
7 0

Answer:

Mass of paraffin that would be needed to collect the amount of energy as 4.73 x 10³ kg of water is 6.85 * 10³ kg

Explanation:

Specific heat capacity of paraffin, Cp = 2.90 J/g.°C = 2900 J/kg.°C

Specific heat capacity of water = 4.20 J/g.°C = 4200 J/kg.°C

Amount of heat energy that can be stored by 4.73 * 10³ kg of water heated through a degree rise in temperature can be calculated as below:

Using  H = mCpθ where m is mass of substance, Cp is specific heat capacity and θ is temperature change

H = 4.73 * 10³ kg * 4200 J/kg.°C * 1°C

H = 19866000 J

Mass of paraffin that would be needed to collect the amount of energy as 4.73 x 10 3 kg of water for a degree rise in temperature is calculated as follows:

H = mCpθ

19866000J = m * 2900 J/kg.°C * 1°C

m = 19866000J / 2900 J/kg

m = 6850.34 kg = 6.85 * 10³ kg

Therefore, mass of paraffin that would be needed to collect the amount of energy as 4.73 x 10³ kg of water is 6.85 * 10³ kg

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Answer:

its iron not m or e

Explanation:

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Answer:

Metallic bonding is found in metals and their alloys. When the atoms give up their valence electrons, they form ions. These ions are held together by the electron cloud surrounding them. Metals are shiny because they have a lot of free (i.e. delocalized) electrons that form a cloud of highly mobile negatively charged electrons on and beneath the smooth metal surface in the ideal case. ... In the absence of any external EM field, the charges in the plasma are uniformly distributed within the metal.

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3 years ago
Which must be the same when comparing 1 mol of oxygen gas, O2, with 1 mol of carbon monoxide gas, CO?
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Answer: C. The Number of Molecules

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6 0
3 years ago
a. Calculate the pH of a solution that has [H3O+]= 4.3 x 10-5 M. b. Is the solution acidic or basic? How do you know?
Luda [366]

Answer:

* pH=4.37

* The solution is acidic since the pH is below 7.

Explanation:

Hello,

In this case, we can mathematically define the pH by:

pH=-log([H_3O^+])

Thus, for the given hydronium concentration we simply compute the pH:

pH=-log(4.3x10^{-5})=4.37

Thereby, we conclude the solution is acidic due to the fact that the pH is below 7 which is the neutral point and above it the solutions are basic.

Regards.

3 0
3 years ago
In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very
SVETLANKA909090 [29]

<u>Answer:</u> The final concentration of potassium nitrate is 5.70\times 10^{-6}M

<u>Explanation:</u>

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2          .......(1)

  • <u>Calculating for first dilution:</u>

M_1\text{ and }V_1 are the molarity and volume of the concentrated KNO_3 solution

M_2\text{ and }V_2 are the molarity and volume of diluted KNO_3 solution

We are given:

M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL

Putting values in equation 1, we get:

7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M

  • <u>Calculating for second dilution:</u>

M_2\text{ and }V_2 are the molarity and volume of the concentrated KNO_3 solution

M_3\text{ and }V_3 are the molarity and volume of diluted KNO_3 solution

We are given:

M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL

Putting values in equation 1, we get:

1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M

Hence, the final concentration of potassium nitrate is 5.70\times 10^{-6}M

8 0
3 years ago
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