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soldier1979 [14.2K]
3 years ago
11

How many atoms are in 123 g of calcium

Chemistry
1 answer:
Keith_Richards [23]3 years ago
5 0

Answer:

Molar mass of Ca = 40 g / mol , given 123 g Ca is 123/40= 3.075 moles,

1 mole = 6.022 * 10^23 atoms, so 3.075 moles Ca= 18.51*10^23 atoms

Explanation:

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What Can bond together to form an ionic bond?
Kazeer [188]

Answer:

An ionic bond forms between two ions of opposite charges. In ionic bonding, electrons transfer from one atom to another. The elements take on either a negative or positive charge. Ions are another name for charged atoms. Some elements are electropositive, and some are electronegative.

Hope this helps!

5 0
3 years ago
how many mL of a 3.5 M sodium hydroxide will be needed to neutralize 15mL of a 4.3 M hydrochloric acid solution?
kobusy [5.1K]
Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O

Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...

Stem 3 : use mole ratio....
HCl : NaOH
1 : 1
So mole is same as calculated above...

Step 4 :
3.5 mole of NaOH is in 1000ml
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6 0
3 years ago
VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu
SOVA2 [1]

Answer:

31.9 °C  

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

q = 12.35 cal

m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹  

ΔT  = 12.35/(2.406 °C⁻¹) = 5.13 °C

(b) Calculate T₂

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

 

6 0
3 years ago
Calculate 5+7*3*<br><br> Your answer:​
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I got the answer 26
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3 years ago
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The octet rule states that, in chemical compounds, atoms tend to have _____.
olga nikolaevna [1]

The correct answer is C


4 0
3 years ago
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