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Natalija [7]
3 years ago
11

How does the temperature and pressure change inside the earth, moving downward from the crust to the core?

Chemistry
2 answers:
Rus_ich [418]3 years ago
8 0

B. Both temperature and pressure increase

Irina-Kira [14]3 years ago
4 0

Answer:

The answer you're looking for is,

   B) Both the temperature and the pressure increase.

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Help me out here please?
Tasya [4]

Answer:

B

Explanation:

3 0
3 years ago
By which mechanism would a steroid molecule diffuse into the cell?
LUCKY_DIMON [66]

Answer: Directly through the phospholipid membrane

Explanation:

The cell membrane consist of a phospholipid bilayers structure. In the interior of the membrane, the phospholipid tail are hydrophobic, which makes the cell membrane to be selectively permeable, it is permeable to non polar molecules and impermeable to polar molecules.

Because Steroids are fat soluble, non polar compounds, they can diffuse directly through the hydrophobic, non polar core of the phospholipid bilayer without the use of carrier proteins.

5 0
3 years ago
Care needs to be taken when adding acid to base because
nikitadnepr [17]
<span>Mixing an acid and a base results in neutralization, but the results are potentially dangerous. No matter which acid or base is used, the resulting solution is water and varying types of salt. The process of neutralization often involves the substances heating up when they come together. If the solution heats up too much or too fast, a violent explosion or the creation of harmful or flammable gases is a possibility. This occurs when the chemicals are mixed too quickly, the acid and base are too strong or if there is no available salt to be made in the solution.</span>
3 0
3 years ago
A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num
-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}

<em>B) Radius</em>

\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}

V = \frac{ 4}{3 }\pi r^{3}

r^{3} = \frac{3V }{4 \pi }\

r= \sqrt [3]{ \frac{3V }{4 \pi } }

r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}

3 0
3 years ago
What is the solubility product expression for Ag2CO3
Dominik [7]
<span> Ksp = [Ag+]^2[CO32-]that should be it </span>
7 0
3 years ago
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