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3 years ago

What is the mass of 6.73x10^23 atoms of carbon

Chemistry

1 answer:

kap26 [50]3 years ago

6 0

Answer: 13.42g

Explanation:

1mole of carbon =12g

1mole (12g) of carbon contains 6.02x10^23 atoms.

Therefore, Xg of carbon will contain 6.73x10^23 atoms i.e

Xg of carbon = (12x6.73x10^23)/6.02x10^23 = 13.42g

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1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.

andriy [413]

Answer:

A. $m_{NH_3}^{theo} =1.50gNH_3$

B. $Y=82.2\%$

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

$N_2+3H_2\rightarrow 2NH_3$

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

$n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\ n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3$

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

$m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3$

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

$Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%$

Best regards!

5 0

3 years ago

PLEASE HELP ME (JOULES)

Svetllana [295]

Answer:

The answer is not 9.9 i used that and got it wrong

Explanation:

8 0

2 years ago

Hydrogen bonds form between neighboring water molecules because of:

12345 [234]

*A & B*

Answers A & B are not possible, as Hydrogen “bonds” are intermolecular forces and do not actually involve transfer or sharing of electrons.

*C & D*

Viscosity and surface tension are not the answer as they are not specific enough to the question.

*E*

Polarity of water molecules is the correct answer, as water molecules are highly polar. The partial positive of the Hydrogen on one water molecule is highly attracted to the partial negative of the Oxygen (due to its lone pairs) on another water molecule.

4 0

3 years ago

The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu

MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation- $k=Ae^{\frac{-E_{a}}{RT}}$ , where k is rate constant, A is pre-exponential factor, $E_{a}$ is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

$\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}$

Here $\frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005}$ , T = 298 K , R = 8.314 J/(mol.K) and $E_{a}^{ala-pro}=60kJ/mol$

So, $\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}$

$\Rightarrow$ $E_{a}^{phe-pro}=65705J/mol=66kJ/mol$ (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

7 0

3 years ago

The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M

anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0

3 years ago

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