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astraxan [27]
3 years ago
11

What is the formula for calcium fluoride​

Chemistry
2 answers:
MatroZZZ [7]3 years ago
7 0

Answer:

CaF_{2}

Explanation:

Calcium = Ca

Fluorine = F

pantera1 [17]3 years ago
6 0

Answer:

CaF2

Explanation:

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Though alchemists were often superstitious, they left a rich legacy for modern chemists. What was their main contribution?
topjm [15]

though alchemists were often superstitious, they left a rich legacy of modern chemists. what was their main contribution-

Explanation:

they were the first to preform experiments.

5 0
3 years ago
The pH scale for acidity is defined by pH=−log10[H⁺] where [H⁺]is the concentration of hydrogen ions measured in moles per liter
masha68 [24]

Answer:

5.62 * 10^-13 moles per liter

Explanation:

The pH of a solution is the negative logarithm to base 10 of the concentration of hydrogen ions. What we simply do here is to input the information in the question into the equation:

pH=−log10[H⁺]

Here we know the pH but we do not know the concentration of the hydrogen ions.

12.25 = -log [H+]

log[H+] = -12.25

[H+] = 10^-12.25

[H+] = 5.62 * 10^-13 moles per liter

7 0
3 years ago
How many moles are present in 25.0 grams of potassium permanganate, KMnO,? ​
poizon [28]

Answer: 0.158

Explanation:

7 0
3 years ago
Which element could be an isotope of the atom?
Anna35 [415]
Element which possess different number of "Neutrons" in different situations could be an isotope of the atom

Ex. - 17 Cl 34, 17 Cl 35, 17 Cl 36

They all the isotopes.

Hope this helps!
4 0
4 years ago
At what temperature would a 1.30 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86
Setler [38]

Answer:

-4.59°C

Explanation:

Let's see the formula for freezing point depression.

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Freezing constant. For water if 1.86°C/m

m = molality (moles of solute in 1kg of solvent)

i = Van't Hoff factor.

0°C - Freezing T° of solution = 1.86°C /m . 1.30m . 1.9

Freezing T° of solution = - (1.86°C /m . 1.30m . 1.9)

Freezing T° of solution = - (1.86°C/m .  1.30m . 1.9) → -4.59°C

NaCl →  Na⁺  +  Cl⁻

4 0
3 years ago
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