All categories

3 years ago

Why would increased levels of greenhouse gases contribute to higher temperatures on earth

1 answer:

5 0

Greenhouse gases are the gases in the atmosphere that keep the air in the atmosphere. It also traps the sun's sunlight. Since the heat is trapped in here, it would contribute to higher temperatures. *Have a nice day!!*

You might be interested in

Consider the reaction below. Which species is(are) the BrÃ¸nsted-Lowry acid(s)?

madreJ [45]

Answer:

HF is the acid

Explanation:

The Brønsted-Lowry theory defines the acids and bases in chemistry as follows:

An acid is the species that can donate a proton

A base can accept protons.

In the reaction:

HF(aq) + NH₃(aq) → NH₄⁺(aq) + F⁻(aq)

As you can see, HF can donate its proton to produce F⁻: HF is the acid

<em>In the same way, NH₃ is accepting a proton, NH₃ is the base.</em>

5 0

3 years ago

You need to produce a buffer solution that has a pH of 5.26. You already have a solution that contains 10. mmol (millimoles) of

Andreas93 [3]

The question is incomplete, complete question is :

You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pka of acetic acid is 4.74.

Answer:

33.11 millimoles of acetate we will need to add to this solution.

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

Where :

tex]pK_a[/tex] = negative logarithm of acid dissociation constant of acid

[salt] = Concentration of salt

[Acid] = Concentration of salt

We have:

pH = 5.26

$pK_a=4.74$

[salt] = $[CH_3COO^-]$ = ?

[acid] = $[CH_3COOH]=10.0 mmol$

$5.26=4.74+\log(\frac{[CH_3COO^-]}{[10.0 mmol]})$

$[CH_3COO^-]=33.11 mmol$

33.11 millimoles of acetate we will need to add to this solution.

3 0

4 years ago

In the following experiments, identify the

Sonja [21]

The tire pressure
Independent variable
Increases as the temperature increases
Dependent variable
hope I helped!

6 0

3 years ago

Read 2 more answers

A helium balloon with a volume of 550mL is cooled from 305 to 265K. The pressure on the gas is reduced from 0.45 atm to 0.25 atm

Aleksandr [31]

860 mL.

<h3>Explanation</h3>

Separate this process into two steps:

Cool the balloon from 305 K to 265 K.
Reduce the pressure on the balloon from 0.45 atm to 0.25 atm.

What would be the volume of the balloon after each step?

After Cooling the balloon at constant pressure:

By Charles's Law, the volume of a gas is directly related to its temperature in degrees Kelvins.

In other words,

$\dfrac{V_2}{V_1} = \dfrac{T_2}{T_1}$ ,

where

$V_1$ and $V_2$ are volumes of the same gas.
$T_1$ and $T_2$ are the temperatures (in degrees Kelvins) of that gas.

Rearranging,

$V_2 = V_1 \cdot \dfrac{T_2}{T_1}\\\phantom{V_2} = 550 \times \dfrac{265}{305}\\\phantom{V_2} = 478 \; \text{mL}$ .

The balloon ended up with a lower temperature. As a result, its volume drops: $V_2 < V_1$ .

After reducing the pressure on the balloon at constant temperature:

By Boyle's Law, the volume of a gas is inversely proportional to the pressure on this gas.

In other words,

$\dfrac{V_2}{V_1} = \dfrac{P_1}{P_2}$ ,

where

$V_1$ and $V_2$ are volumes of the same gas.
$P_1$ and $P_2$ are the pressures on this gas.

Rearranging,

$V_2 = V_1 \cdot \dfrac{P_1}{P_2}\\\phantom{V_2} = 478 \times \dfrac{0.45}{0.25}\\\phantom{V_2} = 860 \;\text{mL}$ .

There's now less pressure on the balloon. As a result, the balloon will gain in volume: $V_2 > V_1$ .

The final volume of the balloon will be $860 \; \text{mL}$ .

7 0

3 years ago

During one researcher's experiment, the total mass of the gas-generating solid and entire assembly was 52.1487g before the react

irinina [24]

Answer:

43.93 g/mol

Explanation:

The mass of the gas before reaction = 52.1487 g

The mass of the gas after reaction = 52.1098 g

Mass of gas generated = 0.0389 g

Moles of the gas = $8.854\times 10^{-4}\ moles$

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Molar\ mass = \frac{Mass\ taken}{Moles}$

$Molar\ mass= \frac{0.0389\ g}{8.854\times 10^{-4}\ moles}$

Molar mass of the gas = 43.93 g/mol

7 0

4 years ago