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ExtremeBDS [4]
3 years ago
8

List in order which digits are significant in the following numbers:

Chemistry
1 answer:
ozzi3 years ago
5 0

Answer:  see below

<u>Explanation:</u>

Zeros between the digit and decimal point are NOT significant.

Zeros between digits ARE significant.

a. 230 mL.              sf: 23

b. 0.00540 s          sf: 540

c. 1440                    sf: 144

d. 50 ℃                  sf: 5

e. 0.9405 g/cm3    sf: 9405

f. 700 jellybeans    sf: 7

g. 1.05 x 10⁶            sf: 105

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What is the temperature of a gas that is expanded from 3.75 L at 37 degrees Celsius to 5.6 L?
deff fn [24]

Answer:

190 °C  

Step-by-step explanation:

The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:  

V₁/T₁ = V₂/T₂      Invert both sides of the equation.  

T₁/V₁ = T₂/V₂      Multiply each side by V₂

T₂ = T₁ × V₂/V₁

=====

V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K  

V₂ = 5.6 L;   T₂ = ?  

=====

T₂ = 310.15 × 5.6/3.75

T₂ = 310.15 × 1.49

T₂ = 463 K

t₂ = 463 – 273.15

t₂ = 190 °C

3 0
3 years ago
One of the commercial uses of sulfuric acid is the production of calcium sulfate and phosphoric acid. If 26.8 g of Ca₃(PO₄)₂ rea
d1i1m1o1n [39]

Answer:

The percent yield reaction is 64.3%

Explanation:

This is the ballanced reaction

Ca₃(PO₄)₂ (s) + 3H₂SO₄ (aq) → 2H₃PO₄ (aq) + 3CaSO₄ (aq)

Let's determine the moles of our reactants:

Mass / Molar mass = Mol

26.8 g / 310.18 g/m = 0.0864 moles of phosphate.

54.3 g / 98.06 g/m = 0.554 moles of sulfuric

1 mol of phosphate reacts with 3 mol of sulfuric so

0.0864 mol of PO₄⁻³ will react with (0.0864 .3)/1 = 0.259 moles

I have 0.554 of sulfuric, so this is the reactant in excess.

The limiting reagent is the Phosphate.

1 mol of PO₄⁻³ produces 2 mol of phosphoric

0.0864 of PO₄⁻³ will produce the double amount (0.0864 .2) = 0.173 moles

Mol . molar mass = Mass

0.173 m . 97.98g/m = 16.95 g (This is the theoretical yield)

Percent yield = (Produced / Theoretical) .100

(10.9 g / 16.95 g) . 100 = 64.3 %

5 0
3 years ago
Read 2 more answers
Why does the titration of a weak acid with a strong base always have a basic equivalence point? Why does the titration of a weak
CaHeK987 [17]

Answer: Option (c) is the correct answer.

Explanation:

When a weak acid reacts with a strong base then it results into the formation of a basic solution. Hence, the resulting solution will always have a pH greater than 7.

Since, at the equivalence point number of hydrogen ions become equal to the hydroxide ions. Therefore, pH of solution will be about 7.

So at the equivalence point, the weak acid will get neutralized due to the addition of strong base. Therefore, it will lead to the formation of conjugate base.

As a result, the solution will become slightly basic in nature.

Thus, we can conclude that at the equivalence point, the acid has all been converted into its conjugate base, resulting in a weakly acidic solution because at the equivalence point, the acid has all been converted into its conjugate base, resulting in a weakly basic solution.

5 0
3 years ago
How many miles of N2o2 in 76 g?molar mass is 92.02g
wlad13 [49]
molar mass of the molecule is 60. Moles=mass/molar mass
92.02g/60=1.53366667mol
1.53mol(3sf)
3 0
3 years ago
Al(OH)3 + 3 HCl = AlCl3 + 3 H2O
Kryger [21]

Answer:

1) 1.235 g.

2) 0.61 g.

Explanation:

  • From the balanced equation:

<em>Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O.</em>

1.0 mol of Al(OH)₃ reacts with 3.0 moles of HCl to produce 1.0 mol of AlCl₃ and 3.0 moles of H₂O.

<em>1) How many grams of HCl can a tablet with 0.880 g of Al(OH)₃ consume? </em>

  • To calculate the amount of HCl needed to consume 0.880 g of Al(OH)₃, we need to calculate the no. of moles of Al(OH)₃:

no. of moles of Al(OH)₃ = mass/molar mass = (0.880 g)/(78.0 g/mol) = 1.13 x 10⁻² mol.

∵ Every 1.0 mol of Al(OH)₃ needs 3.0 moles of HCl to be consumed.

∴ 1.13 x 10⁻² mol of Al(OH)₃ needs (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of HCl.

The no. of grams of HCl = no. of moles of HCl x molar mass of HCl = (3.385 x 10⁻² mol)(36.5 g/mol) = 1.235 g.

<em>2) How much H₂O?</em>

∵ Every 1.0 mol of Al(OH)₃ produces 3.0 moles of H₂O.

∴ 1.13 x 10⁻² mol of Al(OH)₃ produces (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of H₂O.

<em>The no. of grams of H₂O = no. of moles of H₂O x molar mass of H₂O </em>= (3.385 x 10⁻² mol)(18.0 g/mol) = <em>0.6092 g ≅ 0.61 g.</em>

7 0
3 years ago
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