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horrorfan [7]
3 years ago
13

Whats the lewis dot diagram for calcium

Chemistry
1 answer:
stepladder [879]3 years ago
4 0

Draw the Lewis dot structure for each atom of the compound to show how many valence electrons are present in each atom. For example, the calcium atom in calcium chloride, CaCl2, has two valence electrons, and the chlorine atoms have seven valence electrons each.

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If the copper cube had a mass of 28.7 grams and a volume of 3.2 mL, what would its density be?
Otrada [13]

Answer:

ρ=m÷v

 =28.7÷3.2

 ≈8.97 g/m³

3 0
3 years ago
After 11.5 days, 12.5% of a sample of radon-222 that originally weighed 42g remains. What is the half-life of this isotope?
Bond [772]
Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value.  The equation to describe the decay is
Nt=N0(1/2) ^{t/t(1/2)}  where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time.  So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days
5 0
3 years ago
For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ
Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0
2 years ago
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In sublimation which state of matter is absent
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Liquid :) it goes directly from a solid to a gas. Have a nice day!
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