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s2008m [1.1K]
3 years ago
6

(PLEASE HELP ASAP I WILL MARK BRAINLIEST!! C:)Which of the following is a reason why water is so important to the function of ce

lls? Water helps keep the temperature of cells from changing rapidly.
1.Water helps keep the temperature of cells from changing rapidly.

2.Water helps maintain the size and shape of the cell.

4.Without water, most chemical reactions within the cell would not take place. Without water, most chemical reactions within the cell would not take place.

5.All of these answers show how water is important to the function of cells.



!!:(
Chemistry
2 answers:
katrin2010 [14]3 years ago
5 0
The answer is letter 4
liraira [26]3 years ago
4 0
I’d go with choice number 4. Hope this helps!
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Qual a relação e as PROPORÇÕES entre Joules, cal, Cal, e Kcal?
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Answer:

Check Explanation

Verificar explicação

Explanation:

English Translation

What is the relationship and the PROPORTIONS between Joules, lime, Cal, and Kcal?

Joules is the SI unit of Energy (heat energy).

Calorie, also called lime, is the amount of heat required to raise the temperature of 1 gram of pure water by 1°C or 1 K.

1 calorie = 4.186 J

A thousand calorie is equal to 1 kcal.

Hence,

1 kcal = 1000 calories = 4186 J

In Portugese/Em português

Joules é a unidade SI de energia (energia térmica).

Caloria, também chamada de cal, é a quantidade de calor necessária para elevar a temperatura de 1 grama de água pura em 1°C ou 1 K.

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1 kcal = 1000 calorias = 4186 J

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Espero que isto ajude

6 0
3 years ago
An acetate buffer solution is prepared by combining 50. mL of a 0.20 M acetic acid, HC2H3O2 (aq), and 50. mL of 0.20 M sodium ac
Gala2k [10]

Answer:

B. CH3COOH pH > 4.7 (4.8)

Explanation:

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  • CH3COONa + NaOH ↔ CH3COONa

∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol

⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol

⇒ <em>C</em> CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)

⇒ <em>C</em> CH3COOH = 0.0905 M

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⇒ <em>C</em> CH3COONa =  (0.01 mol + 5 E-4 mol) / (0.105 L )

⇒ <em>C</em> CH3COONa = 0.1 M

∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5

⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])

⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]

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