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3 years ago

How does this simulation demonstrate the Law of Universal Gravitation?

Chemistry

2 answers:

Tems11 [23]3 years ago

6 0

It demonstrates this law, by showing the students how mass affects gravity in a solar system.

Hope it helps :)

Verizon [17]3 years ago

5 0

it demonstrates this law by showing the students how mass affects Gravity in a solar system V it is the year 2085 and the world population has grown at an alarming rate.

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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 1.92 g of sodium

Oksana_A [137]

Answer:

The % yield is 27.0 %

Explanation:

Step 1: Data given

Mass of sulfuric acid = 4.9 grams

Mass of sodium hydroxide = 7.8 grams

Mass of sodium sulfate produced = 1.92 grams

Molar mass H2SO4 = 98.08 g/mol

Molar mass NaOH = 40 g/mol

Molar mass Na2SO4 = 142.04 g/mol

Step 2: The balanced equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Step 3: Calculate moles H2SO4

Moles H2SO4 = Mass H2SO4 / Molar mass H2SO4

Moles H2SO4 = 4.9 grams / 98.08 g/mol =

Moles H2SO4 = 0.05 moles

Step 4: Calculate moles NaOH

Moles NaOH = 7.8 grams / 40 g/mol

Moles NaOH = 0.195 moles

Step 5: Calculate limiting reactant

For 1 mole H2SO4 consumed ,we need 2 moles NaOH to produce 1 mole Na2SO4 and 2 moles H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.05 moles).

NaOH is in excess. There will react 2*0.05 = 0.1 moles

There will remain 0.195 -0.1 = 0.095 moles NaOH

Step 6: Calculate moles Na2SO4

For 1 mole H2SO4 consumed ,we need 2 moles NaOH to produce 1 mole Na2SO4

For 0.05 moles H2SO4, we have 0.05 moles Na2SO4

Step 7: Calculate mass of Na2SO4

Mass Na2SO4 = Moles Na2SO4 * Molar mass Na2SO4

Mass = 0.05 moles * 142.04 g/mol = 7.102

This is the theoretical yield

Step 8: Calculate the percent yield of Na2SO4

% yield = (actual yield / theoretical yield) * 100%

% yield = (1.92 / 7.102) *100% = 27.0 %

The % yield is 27.0 %

7 0

4 years ago

Question: What is the molar concentration of 1.29 mol of KCL dissolved in 350 mL of solution?

STatiana [176]

Answer:

M = 3.69 M.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the molar concentration of the 1.29 moles of KCl in 350 mL of solution by recalling the mathematical definition of molarity as the division of the moles by the volume in liters, in this case 0.350 L; thus, we proceed as follows:

$M=\frac{1.29mol}{0.350L}\\\\M=3.69M$

Which gives molar units, M, or just mol/L.

Regards!

8 0

3 years ago

Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse

amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

6 0

3 years ago

Read 2 more answers

How many grams of H20 will be produced by 18g H2

AURORKA [14]

Answer:

160.86 g to the nearest hundredth.

Explanation:

H2 + O ---> H2O

2.016 g H2 produces 18.016 g H2O

1g .. .. .. .. .. 18.016 / 2.016 g H2O

18g .. .. .. .. .. . . (18.016 * 18) / 2.016 g H2O.

= 160.86 g.

8 0

3 years ago

Which linkages in starch and triglycerides result from condensation reactions?

Whitepunk [10]

Answer: The linkages in starch and triglycerides resulting from condensation reactions is glycosidic bonds in starch and ester bonds in triglycerides

8 0

4 years ago