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Nesterboy [21]
3 years ago
8

A bitter liquid ,ph= is classified as

Chemistry
1 answer:
choli [55]3 years ago
8 0
It is a basic liquid. 

Hope it helped!
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Please help me with this it’s a quiz :(
STatiana [176]

Answer:

Carbon dixoide and water combine to form glucose and oxgen.

I hope this helped! :D

Explanation:

this process by which green plants and some other organisms use sunlight to synthesize foods from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a byproduct.

4 0
3 years ago
Why does a planet stay in orbit?
I am Lyosha [343]

Answer:

Gravity

Explanation:

A planet is gravitationally pulled around the sun.

6 0
3 years ago
Read 2 more answers
Half-life is always given in units of time
ad-work [718]
Yes, half life is always given in units of time, hope this helps
5 0
3 years ago
How many grams of KCl are needed to prepare 0.750 L of a 1.50 M solution in water?
dlinn [17]

Answer: Option A) 83.9g

Explanation:

KCl is the chemical formula of potassium chloride.

Given that,

Amount of moles of KCl (n) = ?

Volume of KCl solution (v) = 0.75L

Concentration of KCl solution (c) = 1.5M

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

make n the subject formula

n = c x v

n = 1.5M x 0.75L

n = 1.125 mole

Now given that,

Amount of moles of KCl (n) = 1.125

Mass of KCl in grams = ?

For molar mass of KCl, use the molar masses of:

Potassium, K = 39g;

Chlorine,Cl = 35.5g

KCl = (39g + 35.5g)

= 74.5g/mol

Since, amount of moles = mass in grams / molar mass

1.125 mole = m / 74.5g/mol

m = 1.125 mole x 74.5g/mol

m = 83.81g

Thus, 83.9 grams of KCl are needed to prepare 0.750 L of a 1.50 M solution in water

3 0
2 years ago
What volume will 50.2 grams of co2 (g) occupy at stp?
Genrish500 [490]
Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols

For every 1 mol of gas, there will be
24000 cm^3 of gas

Vol. = 1.14 x 24 dm^3
= 27.36 dm^3
8 0
3 years ago
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