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Zarrin [17]
3 years ago
7

Plants absorb ______ from the atmosphere and release _______ to the atmosphere.

Chemistry
1 answer:
astra-53 [7]3 years ago
6 0
  1. {CO}_{2}
  2. {O}_{2}

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST ☺️

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Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
Many people in the world don't have access to clean drinking water. What may be the cause of this?
Furkat [3]

Answer:

Climate change and an increase in unpredictable and extreme weather is a growing challenge. Long periods of drought affect clean water supplies while flooding can pollute clean water sources and cause outbreaks of disease.

8 0
2 years ago
Read 2 more answers
1. A smoothie contains 1 banana (B), 4 strawberries (St), 1 container of yogurt (Y), and 3 ice cubes (Ic). Write a balanced equa
kvv77 [185]

Answer:

1B +4St+1Y+3lc——-> BSt4Ylc3

Explanation:

I only know the answer for the first question.

4 0
3 years ago
A 50.0g g sample of 16n decays to 12.5g in 14.4 seconds. What is its half life
mixas84 [53]
T is amount after time t 
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>

<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>

<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>

<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>

<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>

<span>HL = 14.4 s / 2 = 7.2 seconds </span>
8 0
3 years ago
Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below.
IceJOKER [234]
2HgO=2Hg + O2
433,18 g. = 32 g
x. = 250 g
x= 3 384,21875 g of HgO is needed
4 0
3 years ago
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