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2 years ago

How does the greenhouse effect relate to the theme of stability and change

1 answer:

8 0

The greenhouse effect increases the temperature of the Earth by trapping heat in our atmosphere. This keeps the temperature of the Earth higher than it would be if direct heating by the Sun was the only source of warming.
When the sunlight reaches the surface of the Earth, some of it is absorbed which warms the ground and some bounces back to space as heat. Greenhouse gases that are in the atmosphere absorb and then redirect some of this heat back towards the Earth.

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It is becoming less expensive to screen blood samples for DNA. Certain diseases have a genetic basis. What is a possible ethical

sveta [45]

Answer:

The answer you would be looking for is option A because all of the other options are either false, or beneficial to us, and i took the test. Thanks

Explanation:

8 0

3 years ago

Read 2 more answers

A period of the periodic table ends with the highest energy level of an element's ators<br> having?"

andrey2020 [161]

Answer:

the last period. i think has the largest energy level

Explanation:

7 0

2 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

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

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

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

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

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

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

3 years ago

Will the solubility of CaSO4 in 0.10 Molar CaF2 be smaller than, larger than, or the same as in pure water?

Kay [80]

Smaller than; less of it will dissolve before the solution is saturated

3 0

3 years ago

Which combination may be used to prepare a buffer having a ph of 8. 8?

7nadin3 [17]

The combination used in the preparation of a buffer with a pH around 8.8 accounts for ka = 7 * 10 -3 for h 3po4

The ph of the buffer can be shown as:

pH = pKa + log [Salt] /[ Acid ]

[Salt] /[ Acid ] = x

For h3po4 with ka= 7 × 10–3

8.8 = - log (7 × 10^–3) + log x

8.8 = 2.21 + log x

Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

For h2po4- with ka= 8 × 10–8

8.8 = - log (8 × 10^–8) + log x

8.8 = 7.14 + log x

Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

For hpo42– with ka= 5 × 10–13

8.8 = - log (5 × 10–13) + log x

8.8 = 12.31 + log x

Thus, the value of log x is coming negative and therefore can not be used for preparing buffer.

Hence, the correct answer is option A

Learn more about buffering systems here,

brainly.com/question/16556401

# SPJ4

4 0

2 years ago