1
3
4
Answer:
2
Explanation:
the pH of the strong acid solution will be
pH = -log[H+] = -log[0.01] = 2
b.The nitrogen and oxygen in Earth's atmosphere keep the surface pleasantly warm.
c.Even in low-Earth orbit, some atmospheric gas is still present.
d.Atmospheric pressure decreases with altitude.
Answer:
b.The nitrogen and oxygen in Earth's atmosphere keep the surface pleasantly warm.
Explanation:
Hello,
Based on the behavior of the Earth's atmosphere, it is widely known that the surface temperature depends on the altitude since the Earth's nucleus has the capacity to warm the outer surface as it is closer to it. In such a way, the warmness of the surface depends on the nucleus rather than the presence of nitrogen and oxygen.
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Among the provided options, the statement falsely attributing the Earth's warmth to nitrogen and oxygen in the atmosphere is incorrect. It is the greenhouse gases that keep Earth's surface warm, not nitrogen and oxygen.
The statement that is not true among the options provided is option b. The nitrogen and oxygen in Earth's atmosphere do not keep the Earth's surface pleasantly warm. It's the greenhouse gases in the atmosphere, which include gases like carbon dioxide and methane, that trap heat from the Sun and keep the Earth's surface warm. Nitrogen and oxygen, which make up the majority of the Earth's atmosphere, do not have this property.
Other statements are correct. Scattering of light by the Earth's atmosphere does cause the blue colour of the sky (a), atmospheric gas is still present even in low-Earth orbit (c), and atmospheric pressure indeed decreases with altitude (d).
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Answer:
There are chemical cross-links between the two strands in DNA, formed by pairs of bases. They always pair up in a particular way, called complementary base pairing: thymine pairs with adenine (T–A) guanine pairs with cytosine (G–C)
Explanation:
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this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is volume of the diluted solution to be prepared
50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL
substituting these values in the formula
1.50 M x 50.0 mL = C x 250 mL
C = 0.300 M
concentration of the final solution is A) 0.300 M
Answer: 0.300M
Explanation:
1) Data:
a) Initial solution
M = 1.50M
V = 50.0 ml = 0.050 l
b) Solvent added = 200 ml = 0.200 l
2) Formula:
Molarity: M = moles of solute / volume of solution is liters
3) Solution:
a) initial solution:
Clearing moles from the molarity formula: moles = M × V
moles of H₂SO₄ = M × V = 1.5M × 0.050 l = 0.075 mol
b) final solution:
i) Volumen of solution = 0.050 l + 0.200l = 0.250l
ii) M = 0.075 mol / 0.250 l = 0.300M ← answeer