describes the forces that exist
between a liquid and different
molecules,
A Adhesion
B. Repulsion
C. Condensation
D. Cohesion
98-2000 Anths Charpentine A Nights Arsenal

Answer:

mass=279grams

Explanation:

GIVEN DATA

number of atoms=3.01×10^23

avogadro's number=6.022×10^22

molar mass of iron=56g/moles

TO FIND

mass in gram of iron=?

SOLUTION

by using the formula

mass in gram=(number of atoms÷avogdro's number)×molar mass

mass=(3.01×10^23÷6.022×10^23)×56

mass=0.499×56

mass=27.9grams=28 grams

Answer:

A.Scientists use genes to explain why evolution happens.

Explanation:

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Answer: The percent yield for the [tex]NaBr[/tex] is, 86.7 %

Explanation : Given,

Moles of [tex]FeBr_3[/tex] = 2.36 mol

Moles of [tex]NaBr[/tex] = 6.14 mol

First we have to calculate the moles of [tex]NaBr[/tex]

The balanced chemical equation is:

[tex]2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr[/tex]

From the reaction, we conclude that

As, 2 moles of [tex]FeBr_3[/tex] react to give 6 moles of [tex]NaBr[/tex]

So, 2.36 moles of [tex]FeBr_3[/tex] react to give [tex]\frac{6}{2}\times 2.36=7.08[/tex] mole of [tex]NaBr[/tex]

Now we have to calculate the percent yield for the [tex]NaBr[/tex].

[tex]\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield = 6.14 moles

Theoretical yield = 7.08 moles

Now put all the given values in this formula, we get:

[tex]\text{Percent yield}=\frac{6.14mol}{7.08mol}\times 100=86.7\%[/tex]

Therefore, the percent yield for the [tex]NaBr[/tex] is, 86.7 %

Answer:

potassium is likely to be harder

Explanation:

because whenan atom loss electron they can form particle ion acquare stable a arrangement that loos electron can transfer to another which that also to acquaire stable

Answer: orbitals supernumerary

Explanation:

Answer:

both S orbital and p orbitals

Explanation:

took the exam got the question right.

Answer:

Radium-226 is a radioactive decay product in the uranium-238 decay series and is the precursor of radon-222. Radium-228 is a radioactive decay product in the thorium-232 decay series. Both isotopes give rise to many additional short-lived radionuclides, resulting in a wide spectrum of alpha, beta and gamma radiations.

Answer: Thus the yield of uranium from 2.50 kg [tex]U_3O_8[/tex] is 2.12 kg

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number [tex](6.023\times 10^{23})[/tex] of particles.

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}[/tex]

moles of [tex]U_3O_8=\frac{2.50\times 1000g}{842g/mol}=2.97mol[/tex]    (1kg=1000g)

As 1 mole of [tex]U_3O_8[/tex] contains = 3 moles of U

2.97 mole of [tex]U_3O_8[/tex] contains = [tex]\frac{3}{1}\times 2.97=8.91moles[/tex] moles of U

Mass of Uranium=[tex]moles\times {\text {Molar mass}}=8.91mol\times 238g/mol=2120g=2.12kg[/tex]  

 ( 1kg=1000g)

Thus the yield of uranium from 2.50 kg [tex]U_3O_8[/tex] is 2.12 kg

Answer:

Rate of hydrogen formation is 0.05 mole per second

Explanation:

Firstly, we write the equation of reaction.

When alkali earth metals react with dilute mineral acid, the reaction is vigorous with the production or evolution of hydrogen gas as a result of the displacement of the hydrogen from the acid by the metal. This is one of the basic reactions of mineral acids

Ca + H2SO4 ——> CaSO4 + H2

Looking at the reaction, 1 mole of calcium gave 1 mole of the hydrogen gas

What we do now is to calculate the number of moles of calcium produced by 20g of Ca

Mathematically;

number of moles = mass/atomic mass

number of moles of calcium is thus

20/40 = 0.5 moles

Now, if 1 mole of calcium produced 1 mole of the gas

Definitely, 0.5 mole of calcium will produce 0.5 mole of the gas

So the rate of gas formation would be 0.5/10 = 0.05 mole/second

Answer:

ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

(1) C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

(2) CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

(4) C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

(5) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = -747.54kJ

Thus ΔH° of the reaction is -747.54kJ

ΔH° of the reaction is for the production of SNG -747.54kJ

According to Hess’ law of constant summation, the standard reaction enthalpy is independent of the pathway or number of the reaction steps taken for a multistep reaction, rather it is the sum of standard enthalpies of intermediate reactions involved at the same temperature.

Based on Hess law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

From the given reactions:

The sum of Reaction 4 × 4 + Reaction 5 - Reaction 1 × 4 gives the reaction below:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = {-74.81 kJ × 4} - 890.3 kJ {- 4 ×-110.5}

ΔH° = -747.54kJ

Therefore, ΔH° of the reaction is for the production of SNG -747.54kJ

Learn more about enthalpy and Hess' law at: https://brainly.com/question/9328637

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate