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Inga [223]
3 years ago
6

I NEED HELP PLS. I WILL GIVE 10 POINTS PLEASEEEEEEES

Chemistry
2 answers:
qwelly [4]3 years ago
6 0

Answer:

Catalyzes

Explanation:

Airida [17]3 years ago
5 0

Answer:

Explanation:

It is the fourth one, catalyzing.

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How many grams of calcium chloride are dissolved in 5.65 liters of a 0.11 m solution of calcium choride?
Julli [10]
C = 0.11 mol
V = 5.65 L
n = ???

n = C*V
n = 0.11 * 5.65
n = 0.622 mols

1 mol of CaCl2 = 40 + 2*35.5 = 111 grams
0.622 mol = x

x = 111 * 0.622
x = 69.0 grams CaCl2
4 0
3 years ago
The measurement 3.8 x 10^3 g could also be written as
solmaris [256]

Right now it's written in scientific notation, so you can just move the decimal place in 3.8 to the right 3 times (as it is times 10 to the third power) to get 3,800g.

5 0
3 years ago
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What element has an atomic mass 4?
Daniel [21]

Answer:

Helium has atomic mass  4.002602.

So the answer to yr question is helium.

4 0
3 years ago
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Butanoic acid + 2 propanol
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Butanoic acid and 2-propanol reacts to form isopropyl butyrate.

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brainliest plz

4 0
2 years ago
Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.
Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0
3 years ago
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