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2 years ago

What would happen to the temreture if there was a sudden icrease in the amount of greenhouse gasses

Chemistry

1 answer:

Rus_ich [418]2 years ago

4 0

Answer:

The higher concentrations of greenhouse gases—and carbon dioxide in particular—is causing extra heat to be trapped and global temperatures to rise.

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. HELPPP 10 PTS WILL MARK BRAINLEST Explain: A student wants to increase the maximum amount of sugar that can dissolve in water.

Dovator [93]

When the student crushes sugar there there occurs an increase in the number of molecules or particles of sugar. This means that now more number of sugar particles (solute) are able to interact with water (solvent).
Therefore, there will be more number of collisions taking place between the solute and solvent molecules.
As a result, sugar will dissolve readily into water as there is an increase in interaction between solute and solvent particles.
Thus, we can conclude that when he crushes the sugar and then stirs it into the water then it will work as this leads to more interaction or collisions between the solute and solvent particles.

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3 years ago

What are the three ways that an object can interact with a static field?

Kitty [74]

Answer:

friction, conduction, and induction

Explanation:

4 0

3 years ago

Identify the components that must be on a data table. *

jolli1 [7]

Answer:

OwO uhm where do you need help with?

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3 years ago

Question 6 Simplify the following expression -3 + 4(a - 3b + 2) - 514 - a + 3b)

Nimfa-mama [501]

This expression simplified is 3a-9b-509

3 0

3 years ago

A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to

grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

Step 1: Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

Step 2: Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

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

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0

4 years ago