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Vinvika [58]
3 years ago
14

Applying the rules of significant figures, which of the following would be the correct value for the density of a substance whic

h has a mass if 32.40 g and a volume of 2.0mL?
16.20 g/mL
16 g/mL
0.06 g/mL
5.00 g/mL​
Chemistry
1 answer:
lilavasa [31]3 years ago
6 0

16 g/ml

<h3>Further explanation</h3>

Given

mass = 32.40 g

volume = 2.0mL

Required

Density

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

ρ = density  

m = mass  

v = volume  

Input the value :

ρ = 32.40 g / 2.0 ml

ρ = 16.2 g/ml

Rules for division: The least number of significant figures in the problem determines the number of significant figures in the answer.

32.40 = 4 sig.fig

2.0 = 2 sig fig

The answer must be 2 sig fig

So the answer = 16 g/ml

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Answer:

33.30 grams of CaCl2 will be required

Explanation:

Given,

Volume of solution, V= 250 ml

Molarity of solution, M= 1.20 mol/L

Molecular mass of CaCL2, S= 40+(35.5 X 2)= 111

We know,

Required mass, W= SVM/1000

Now,

W = (111 X 250 X 1.20)/1000

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Therefore, 33.30 grams of Calcium Chloride will be required.

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2 years ago
Ca(OH)2 + H3PO4 = Ca3(PO4)2 + H2O
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Make a quick chart with each element represented, and count them up. HINT - leave the polyatomic anions together - in this case, PO4

Left      Right
1     Ca  3
2      O   1
5      H   2
1    PO4 2

Begin by balancing like finding common denominators of fractions - apply to both sides:
I started by adding a 2 in front of H3PO4 on the left, them 6 in front of H2O on the right. Last, a 3 in front of Ca (OH)2. Then, re-count using the chart format to make sure you're right.

3Ca(OH)2 + 2H3PO4 = Ca3(PO4)2 + 6H2O

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Read 2 more answers
6.0 mol NaOH reacts with
lina2011 [118]

Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • NaOH: 3 moles
  • H₃PO₄: 1 mole
  • H₂O: 3 moles
  • Na₃PO₄: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?

moles of NaOH=\frac{9 moles of H_{3} PO_{4} x3 moles of NaOH}{1 mole of H_{3} PO_{4}}

moles of NaOH= 27 moles

But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.

<h3>Moles of Na₃PO₄ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?

moles of Na_{3}P O_{4} =\frac{6  moles of NaOHx1 mole of Na_{3}P O_{4} }{3 moles of NaOH}

<u><em>moles of Na₃PO₄= 2 moles</em></u>

Then, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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Answer:

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