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serg [7]
3 years ago
8

Plants that have thick, fleshy stems and waxy leaves to prevent water loss are known as ________ plants.

Chemistry
1 answer:
Mrac [35]3 years ago
5 0
Succulent or a cactus
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I really need help with this before 5:10
gtnhenbr [62]
CAN YOU HELP ME ANWSER MINE ILL HELP YOU
4 0
3 years ago
How many atoms are in 6.30 moles of sulfur (S)?
STatiana [176]

Answer: 3.79*10^24 atoms

Explanation:

1 mole = 6.02214076*10^23 atoms

8 0
3 years ago
When two gases or two liquids form a solution, the substance that is present in the largest amount is the?
skad [1K]
Is the solvent (Hope this helped)
4 0
3 years ago
A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3
lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where  [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio  [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2  can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA]  = 4.63 - 4.20 =  log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 =  [A⁻]/[HA] = 2.69

 [A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of  C6H5CO2⁻ the student needs to dissolve  in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L  = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0
3 years ago
What is in a microscope the enlarges things so our eye can see them?
Fittoniya [83]

Answer:lenses

Explanation:

5 0
3 years ago
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