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zavuch27 [327]
2 years ago
14

How many moles are there in 7.24 grams of calcium carbonate? With work shown

Chemistry
2 answers:
balu736 [363]2 years ago
8 0

Answer:

0.0723371390261859

slava [35]2 years ago
5 0
CaCO3 = 40 + 12 + 16*3 = 100 g/mol
Mol CaCO3 = m/M = 7.24/ 100 = 0.0724 mol
Can you mark it brainliest?
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If an object that stands 3 centimeters high is placed 12 centimeters in front of a plane mirror, how far from the mirror is the
ryzh [129]

Answer:

the mirror is 12 cm away from the image

Explanation:

; the image will be virtual and it will form at 12 cm distance behind the mirror.

7 0
3 years ago
Acidic solutions contain high concentrations of
finlep [7]

Answer:

hydrogen ions

Explanation:

because acid is the specie that have ability to donate proton or forming bond with electron pair

6 0
3 years ago
What are 2 of the characteristics of temporary physical change
Natali5045456 [20]

Answer:

Temporary in nature.

No new substance is formed.

Explanation:

Temporary in nature: Does not affect the internal structure of a substance, only the molecules are rearranged.

No new substance is formed: Most of the physical changes are reversible. We can obtain the substance back even after the change.

hope this helps

have an awesome day -TJ

8 0
2 years ago
Read 2 more answers
Which Bond is the Least Polar? <br><br> a) H-O <br> b) H-F<br> c) H-I<br> d) H-H
Elden [556K]
I'm not sure on this I'm sorry I can't help you I wish I could!
7 0
3 years ago
A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
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