Calculate the [H+] for a solution with a pOH of 4.75

What is the original source of energy that is stored in the food that animals eat?

erastova [34]

Since plants turn sunlight into glucose (sugar) that humans and animals eat... the energy we get from our food comes indirectly from the sunlight.

5 0

3 years ago

How to seperate sand and water

strojnjashka [21]

Answer: Some kind of strainer system

Explanation:

8 0

3 years ago

Read 2 more answers

A lake has been infected by some type of new algae that is unknown. Every single day the amount of surface area that the algae t

FrozenT [24]

Answer:

It takes 86 days take to cover half of the lake

Explanation:

In the day #1, the amount of the algae is X,

In the day #2 is 2X

In the day #3 is 2*2*X = X*2²

...

In the day #n the amount of the algae is X*2^(n-1)

Assuming X = 1m³. In the day 87, the area infected was:

1m³*2^(87-1)

7.74x10²⁵m³ is the total area of the lake

the half of this amount is 3.87x10²⁵m³

The time transcurred is:

3.87x10²⁵m³ = 1m³*2^(n-1)

Multiplying for 5 in each side:

ln (3.87x10²⁵) = ln (2^(n-1))

58.9175 = n-1 * 0.6931

85 = n-1

86 = n

<h3>It takes 86 days take to cover half of the lake</h3>

4 0

2 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

3 years ago

Choose all the answers that apply.

vovangra [49]

Diabetes, ALS, and Alzheimeir's.

If this was the appropriate answer make sure to mark as the brainliest!
-procklown

7 0

3 years ago