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3 years ago

Would you be able to fit a mole of rice in a 30.0m8.5m3.5m room??

Chemistry

1 answer:

Julli [10]3 years ago

8 0

Answer:

No.

Explanation:

1. Volume of a rice grain

There is no standard for the size of a rice grain, so let's make an arbitrary assumption.

An average rice grain behaves as if it were a rectangular solid with dimensions 7 mm × 2 mm × 2 mm.

The volume of one rice grain is

V = lwh = 7 mm × 2 mm × 2 mm = 28 mm³

Convert to cubic metres :

$V = \text{28 mm}^{3} \times \left (\dfrac{\text{1 m}}{\text{1000 mm}}\right )^{3} = \mathbf{2.8 \times 10^{-8}}\textbf{ m}^{\mathbf{3}}$

2. Volume of the room

V = lwh = 30.0 m × 8.5 m × 3.5 m = 890 m³

3. Volume of a mole of rice

$V= 6.022 \times 10^{23}\text{ grains} \times \dfrac{2.8 \times 10^{-8}\text{ m}^{3}}{\text{1 grain }} = \mathbf{1.7 \times 10^{16}}\textbf{ m}^{\mathbf{3}}$

4. Conclusion

The room is not big enough to hold a mole of rice.

If you work it out, you will find that it takes about

20 000 000 000 000 (twenty trillion) rooms to hold a mole of rice.

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Yes. Stars use fusion to create nuclear energy, which is what makes them "alive". The older they are, the "bigger" the element in them is. Hydrogen turns into Helium, and when hydrogen is used up, the helium starts fusing into bigger elements. it stops at iron however. Once stars start fusing silicon to iron, it is doomed because it takes more energy than it gives off.

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3 years ago

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

3 years ago

Push your palms again a wall. What happened

kondaur [170]

Answer:

It felt good!!!

Why do you ask??

3 0

3 years ago

An analytical chemist weighs out 0.093g of an unknown monoprotic acid into a 250mL volumetric flask and dilutes to the mark with

Kamila [148]

Answer:

The molar mass of the unknown acid is 89 g/mol

Explanation:

Step 1: The balanced equation

HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

Step 2: Data given

Mass of the acid = 0.093 grams

volume = 250 mL

titrates with 0.16 M NaOH

adds 6.5 mL NaOH

Step 3: Calculate moles of NaOH

We know the concentration and volume of NaOH needed to neutralize the acid.

By determining the moles of NaOH in that volume in liters (95.9mL=0.0959L), the moles of acid in the original sample can be determined from the reaction stoichiometry.

Moles = Molarity * Volume

Moles = 0.16 M * 0.0065 L

Moles = 0.00104 moles NaOH

Step 4: Calculate moles of the unknown acid:

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

For 0.00104 moles NaOH we have 0.00104 moles of HA

Step 5: Calculate the molar mass of the acid

Molar mass Ha = Mass Ha / moles Ha

Molar mass Ha = 0.093 grams / 0.00104 moles

Molar mass Ha = 89.42 g/mol ≈89 g/mol

The molar mass of the unknown acid is 89 g/mol

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3 years ago

A children's liquid cold medicine has a specific gravity of 1.23. If a child is to take 1.5 tsp in a dose, what is the mass (in

julia-pushkina [17]

Assume 1 tsp is approximately can hold 5 mL liquid.

Given the dose of medicine = 1.5 tsp

Converting 1.5 tsp to mL:

$1.5 tsp * \frac{5 mL}{1 tsp}$ = 7.5 mL

Given the specific gravity of the medicine = 1.23

That means density of the medicine with respect to water will be 1.23

As the density of water is 1 g/mL

We can take density of the medicine to be 1.23 g/mL

Calculating the mass of medicine in grams:

$7.5 mL * \frac{1.23 g}{mL} =9.225 g$

9.225 g medicine is present in one dose.

3 0

3 years ago

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Would you be able to fit a mole of rice in a 30.0m*8.5m*3.5m room??

Would you be able to fit a mole of rice in a 30.0m8.5m3.5m room??