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3 years ago

What are the 3 best solutions to reducing or stopping the effects of climate change?

Chemistry

2 answers:

Anastasy [175]3 years ago

5 0

Answer:

1. reduce green house emissions

2. remove carbon from atmosphere

3. governmental policies

Explanation:

1. the green house gases form like a blanket absorbing radiation and not not letting it escape, gradually heating up the earths surface. so a reduction on energy use is needed.

2. this can be done by planting more trees or plants in general

3. governmental policies are needed to regulate green house gases and limit them.

postnew [5]3 years ago

4 0

Answer:

1. Stop Cutting down Trees

2. Control Population

3. Less use of Fossil Fuels

Explanation:

your welcome :)

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Which sample is most likely to experience the smallest temperature change upon observing 55KJ of heat?

Zigmanuir [339]

Answer:

100 g of water: specific heat of water 4.18 J/g°C

Explanation:

To know the correct answer to the question, we shall determine the temperature change in each case.

For 100 g of water:

Mass (M) = 100 g

Specific heat capacity (C) = 4.18 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 100 x 4.18 x ΔT

Divide both side by 100 x 4.18

ΔT = 55000/ (100 x 4.18)

ΔT = 131.6 °C

Therefore the temperature change is 131.6 °C

For 50 g of water:

Mass (M) = 50 g

Specific heat capacity (C) = 4.18 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 50 x 4.18 x ΔT

Divide both side by 50 x 4.18

ΔT = 55000/ (50 x 4.18)

ΔT = 263.2 °C

Therefore the temperature change is 263.2 °C

For 50 g of lead:

Mass (M) = 50 g

Specific heat capacity (C) = 0.128 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 50 x 0.128 x ΔT

Divide both side by 50 x 0.128

ΔT = 55000/ (50 x 0.128)

ΔT = 8593.8 °C

Therefore the temperature change is 8593.8 °C.

For 100 g of iron:

Mass (M) = 100 g

Specific heat capacity (C) = 0.449 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 100 x 0.449 x ΔT

Divide both side by 100 x 0.449

ΔT = 55000/ (100 x 0.449)

ΔT = 1224.9 °C

Therefore the temperature change is 1224.9 °C.

The table below gives the summary of the temperature change of each substance:

Mass >>> Substance >> Temp. Change

100 g >>> Water >>>>>> 131.6 °C

50 g >>>> Water >>>>>> 263.2 °C

50 g >>>> Lead >>>>>>> 8593.8 °C

100 g >>> Iron >>>>>>>> 1224.9 °C

From the table given above we can see that 100 g of water has the smallest temperature change.

5 0

3 years ago

Which graph correctly shows the effect on the freezing point caused by increasing the molality of a solution ?

beks73 [17]

The answer to your question is graph A :)

6 0

3 years ago

Read 2 more answers

Isotopes of elements have different:

Morgarella [4.7K]

Isotopes of elements have same atomic numbers and different mass number(atomic masses).

7 0

3 years ago

When a car uses a gallon of gas, how much carbon dioxide is emitted into the environment? I give branlyest

sineoko [7]

Answer:

19.5

Please tell me if wrong.

7 0

3 years ago

Read 2 more answers

A quantity of N2 gas originally held at 5.23 atm pressure in a 1.20 −L container at 26 ∘C is transferred to a 14.5 −L container

lara31 [8.8K]

Answer:

n (N₂) = 0.256 mol

n (O₂) = 1.0848 mol

n (Total) = 1.3408 mol

Pressure in new Container = 2.222 atm

Explanation:

Data Given:

For Nitrogen gas (N₂)

Pressure of N₂ gas = 5.23 atm

Volume of N₂ gas = 1.20 L

Temperature of N₂ gas = 26° C

Temperature of N₂ gas in Kelven (K) = 26° C +273

Temperature of N₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas N₂ gas = ?

For Oxygen gas (O₂)

Pressure of O₂ gas = 5.21 atm

Volume of O₂ gas = 5.21 L

Temperature of O₂ gas = 26° C

Temperature of O₂ gas in Kelven (K) = 26° C +273

Temperature of O₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas O₂ gas = ?

*we also have to find the total Pressure in the new container = ?

Formula Used

PV =nRT

n (N₂) = PV /RT . . . . . . . . . . . . . (1)

* Find the quantity of N₂

Put value in formula (1)

n (N₂) = 5.23 atm x 1.20 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

n (N₂) = 6.276 atm .L / 24.52 L atm. mol⁻¹ x 299K

n (N₂) = 0.256 mol

* Find the quantity of O₂

Put value in formula (1)

n (O₂) = 5.21 atm x 5.10 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

n (O₂) = 26.6 atm .L / 24.52 L atm. mol⁻¹

n (O₂) = 1.0848 mol

*Now to find the Total Quantity of both gases

n(Total) = n (N₂) + n (O₂)

n (Total) = 0.256 mol + 1.0848 mol

n (Total) = 1.3408 mol

**To find the Total Pressure in the new Container

Data to calculate Total Pressure in new container

Volume of gas = 14.5 L

Temperature of gases = 20° C

Temperature of gases in Kelven (K) = 20° C +273

Temperature of gases in Kelven (K) = 293K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

Volume Pressure in new container = ?

Formula Used

PV =nRT

P = nRT / V . . . . . . . . . . . . . (2)

Put values in Equation (2)

P = 1.3408 mol x 0.08206 L atm K⁻¹ mol⁻¹ x 293 K / 14.5 L

P = 2.222 atm

8 0

3 years ago