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ipn [44]
4 years ago
12

How many moles are in 74 g KI?

Chemistry
1 answer:
nexus9112 [7]4 years ago
7 0
N=?
m=74g
n=m:Mr
Mr (Cl)=35,5g/mol
n=74g:35,5g/mol
n=2,08mol
I think this is the answer, hope I am right!
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Calculate the pH and pOH of 500.0 mL of a phosphate solution that is 0.285 M HPO42– and 0.285 M PO43–. (Ka for HPO42- = 4.2x10-1
jekas [21]

Answer: when concentrations of acid and base are same, pH = pKa

PH = 12.38 pOH = 1.62

Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00

8 0
3 years ago
25 POINTS
Rudiy27

EASY AS PIE AND I LIKE PIE

Calcium iodide (CaI2) is an ionic bond, which means that electrons are transferred.  In order for Ca to become the ion Ca2+, the calcium atom must lose 2 electrons. (Electrons have a negative charge, so when an atom loses 2 electrons, its ion becomes more positive.)  In order for I to become the ion I1−, the iodine atom must gain 1 electron. (When an atom gains an electron, its ion will be more negative.)  However, the formula for calcium iodide is CaI2 - there are 2 iodine ions present. This makes sense because the iodine ion has a charge of -1, so two iodine ions have to be present to cancel out the +2 charge of the calcium ion.  Therefore, the calcium atom transfers 2 valence electrons, one to each iodine atom, to form the ionic bond.

IF WRONG, SORRY

8 0
4 years ago
Read 2 more answers
A buffer is formed by mixing 100 ml of 0.1 m nh4cl with 50 ml of 0.3 m nh3. to this solution 13 ml of 0.1 m naoh is added. what
Hoochie [10]
I had the same question last week and my answer was C. Try that.
8 0
4 years ago
Can someone help me with this
Trava [24]

Answer:

I think cold front if not than its c

7 0
3 years ago
50.0 ml of 0.010m naoh was titrated with 0.50m hcl using a dropper pipet. if the average drop from the pipet has a volume of 0.0
creativ13 [48]

25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.

The equation of the reaction is;

NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)

We can use the titration formula;

CAVA/CBVB = NA/NB

CA= concentration of acid

VA = volume of acid

CB = concentration of base

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

CB = 0.010 M

VB = 50.0 ml

CA = 0.50 M

VA = ?

NA = 1

NB = 1

Substituting values;

CAVANB = CBVBNA

VA =  0.010 ×  50.0 × 1/ 0.50 × 1

VA = 1 ml

Since the total volume of acid used is 1 ml and each drop contains 0.040 ml

The number of drops required is 1ml/0.040 ml = 25 drops

Learn more: brainly.com/question/1527403

4 0
3 years ago
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