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1 year ago

Energy and Temperature Activity Worksheet

1 answer:

5 0

Land surfaces are much darker so they absorb much more solar radiation than water. Most of the solar radiation is reflected by water which leads in reflecting the solar radiation that reaches the surface back in to the atmosphere.

How does this occur?

Due to lower specific heat of sand it heats up faster than water. On the other hand water requires one calorie of energy to raise the temperature of one gram of water by one degree Celsius. Sand takes 19 calories per gram to raise the temperature of the sand by 1 degree Celsius. While Sand requires fewer calories to raise its temperature one degree Celsius and that’s why it heats up faster than water.

Land absorbs more solar radiation the land surface retains more heat as do the vegetation for energy.

Therefore, during this experiment the sand heats up much faster than water. This happens because of the specific heat of the sand that initiates the heating rapidly under the lamp as the time passes.

Learn more about energy and temperature activity from the link given below.

brainly.com/question/28963012

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How many grams of Fe2+ are there in 0.003109 moles of Fe2+?

Sergio [31]

Answer: 0.174 g

Explanation:

First of all understand that charge on an atom like here 2+ will not vary molar mass. because ions are formed by loss or gain of electrons and electrons do not contribute to the mass of an atom or ion. so losing or gaining electron doesnt create any difference.

molar mass of Fe, i.e. mass of 1 mole of Fe is 56 g. So, mass of 1 mole of Fe2+ will also be 56 g.

given is 0.003109 moles of Fe2+,

Use maths,

1 mole of Fe2+ weighs 56 g

So, 0.003109 moles will weigh = (56 g / 1 mol ) x 0.003109 = 0.174 g.

7 0

2 years ago

What happens if you try to move the atoms very close to each other?

stellarik [79]

They push away from each other or repel due to the same charge they have.

4 0

2 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

Boiling point of a solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water. kB= 0.512 c/m.

Nookie1986 [14]

100.133 degree celsius is the boiling point of the solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water.

Explanation:

Balanced eaquation for the reaction

CaCl2 + 2H20 ⇒ Ca(OH)2 + HCl

given:

mass of CaCl2 = 15.2 grams

mass of the solution = 57 grams

Kb (molal elevation constant) = 0.512 c/m

i = vont hoff factor is 1 as 1 mole of the substance is given as product.

Molality is calculated as:

molality = $\frac{grams of solute}{grams of solution}$

= $\frac{15.2}{57}$

= 0.26 M

Boiling point is calculated as:

ΔT = i x Kb x M

= 1 x 0.512 x 0.26

= 0.133 degrees

The boiling point of the solution will be:

100 degrees + 0.133 degrees (100 degrees is the boiling point of water)

= 100.133 degree celcius is the boiling point of mixture formed.

4 0

3 years ago

What will be the final temperature of the solution in a coffee cup calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is added

padilas [110]

Answer:

21.21°C will be the final temperature of the solution in a coffee cup calorimeter.

Explanation:

$HCl+NaOH\rightarrow H_2O+NaCl$

$\Delta H$ = enthalpy change = -57.2 kJ/mol of NaOH

Moles of sodium hydroxide = n

Molarity of the NaOH = 0.250 M

Volume of NaOH solution = V = 50.00 mL = 0.050 L

$n=Molarity\times V=0.250 M\times 0.050 L= 0.0125 mol$

Moles of HCl = n'

Molarity of the HCl= 0.250 M

Volume of HCl solution = V' = 50.00 mL = 0.050 L

$n'=Molarity\times V=0.250 M\times 0.050 L= 0.0125 mol$

Since 1 mole of Hcl reacts with 1 mole of NaoH. Then 0.0125 mole of HCl will react with 0.0125 mole of NaOH.

The enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

$q=\Delta H\times n=-57.2 kJ/mol \times 0.0125 mol= -0.715 kJ=-715 J$

q = heat released on reaction= -715 J

now, we calculate the heat gained by the solution.:

Q= -q = -(-715 J) = 715 J

m = mass of the solution = ?

Volume of the solution formed by mixing, v = 50.00 ml + 50.00 mL = 100.00 mL

Density of the solution = density of water = d = 1 g/mL

$mass=density\times volume=d\times v=1 g/ml \times 100.00 ml=100 g$

m = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{f}$ = final temperature = ?

$T_{i}$ = initial temperature = $19.50^oC$

$Q=mc\times (T_{f}-T_{i})$

Now put all the given values in the above formula, we get:

$715 J=100 g\times 4.18J/^oC\times (T_f-19.50)^oC$

$T_f=21.21 ^oC$

21.21°C will be the final temperature of the solution in a coffee cup calorimeter.

5 0

3 years ago