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2 years ago

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The tiara worn by Kate Middleton for her wedding to Prince William of England contains 888 diamonds and belongs to the British m

onarchy. If each diamond in the tiara is 1.0 carat, and given that diamond is a form of carbon and that 1 carat is defined as 0.200 g, calculate the number of atoms in the gemstones of that tiara.

1 answer:

allsm [11]2 years ago

5 0

Answer:

The number of atoms in the gemstones of that tiara is $8.9125\times 10^{24} atoms$ .

Explanation:

Number of diamonds ion tiara = 888

Mass of each diamond = 1.0 carat = 0.200 g (given)

Mass of 888 diamonds in tiara:

$888\times 0.200 g=177.600 g$

Given that diamond is a form of carbon.

Atomic mass of carbon atom = 12 g/mol

Moles of 77.600 g of carbon = $\frac{177.600 g}{12 g/mol}=14.800 mol$

Number of atoms of carbon 14.800 moles:

$14.800 mol\times 6.022\times 10^{23} atoms =8.9125\times 10^{24} atoms$

The number of atoms in the gemstones of that tiara is $8.9125\times 10^{24} atoms$ .

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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

2 years ago

A chemist has 2.0 mol of methanol (CH3OH). The molar mass of methanol is 32.0 g/mol. What is the mass, in grams, of the sample?

rodikova [14]

Answer:

$\boxed {\boxed {\sf D. \ 64 \ grams }}$

Explanation:

Given the moles, we are asked to find the mass of a sample.

We know that the molar mass of methanol is 32.0 grams per mole. We can use this number as a fraction or ratio.

$\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}$

Multiply by the given number of moles, which is 2.0

$2.0 \ mol \ CH_3OH *\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}$

The moles of methanol will cancel each other out.

$2.0 \ *\frac{32 \ g \ CH_3OH}{1 }$

The denominator of 1 can be ignored.

$2.0 * 32 \ g\ CH_3OH$

Multiply.

$64 \ g \ CH_3OH$

There are 64 grams of methanol in the sample.

3 0

2 years ago

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

6 0

2 years ago

If 4.9 kg of CO2 are produced during a combustion reaction, how many molecules of CO2 would be produced?

solmaris [256]

Answer:

6.7 x 10²⁶molecules

Explanation:

Given parameters

Mass of CO₂ = 4.9kg = 4900g

Unknown:

Number of molecules = ?

Solution:

To find the number of molecules, we need to find the number of moles first.

Number of moles = $\frac{mass}{molar mass}$

Molar mass of CO₂ = 12 + 2(16) = 44g/mol

Number of moles = $\frac{4900}{44}$ = 111.36mole

A mole of substance is the quantity of substance that contains the avogadro's number of particles.

1 mole = 6.02 x 10²³molecules

111.36 moles = 111.36 x 6.02 x 10²³molecules = 6.7 x 10²⁶molecules

5 0

2 years ago