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2 years ago

What percentage of the mass of a carbon-12 atom is

Chemistry

1 answer:

Lynna [10]2 years ago

7 0

Answer:

% of mass of electrons in C = 0.0272

Explanation:

Atomic no. of carbon = 6

So, no. of electron = 6

Mass of an electron = $\frac{1}{1836} amu$

Mass of 6 electrons = $6 \times \frac{1}{1836} = 0.003268 amu$

Mass of C = 12.0107 u

% of mass of electrons in C = $\frac{Mass\ of\ all\ electrons}{Mass\ of\ C}\times 100$

% of mass of electrons in C = $\frac{0.003268}{12.0107}\times 100=0.0272 \%$

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In animals, nitrogenous wastes are produced mostly from the catabolism of _____. In animals, nitrogenous wastes are produced mos

RoseWind [281]

Answer:

Proteins and nucleic acids

Explanation:

Nitrogen compounds in animals that are no longer of use, or are in access are excreted from the animals body, and are thus called nitrogenous waste. These nitrogenous waste can be excreted in three different ways.

1. Ammonia

2. Urea

3. Uric acid

8 0

2 years ago

What type of ions do nonmetals naturally form?

tensa zangetsu [6.8K]

Non-metal atoms gain an electron, or electrons, from another atom to become >negatively charged ions.

8 0

2 years ago

Read 2 more answers

A cylinder with a moving piston expands from an initial volume of 0.350 L against an external pressure of 1.90 atm. The expansio

myrzilka [38]

Answer:

1.84 L

Explanation:

Using the equation for reversible work:

$W = -P*(V_{2} - V_{1})$

Where:

W is the work done (J) = -287 J.

Since the gas did work, therefore W is negative.

P is the pressure in atm = 1.90 atm.

However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)

Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J

$V_{1}$ is the initial volume = 0.350 L

$V_{2}$ is the final volume = ?

Thus:

(-287 J)*0.00987 (L*atm)/J = -1.9 atm*( $V_{2}$ - 0.350) L

$V_{2}$ = [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L

7 0

2 years ago

I want to know the steps.

Artyom0805 [142]

The answer for the following problem is described below.

Therefore the standard enthalpy of combustion is -2800 kJ

Explanation:

Given:

enthalpy of combustion of glucose(Δ $H_{f}$ of $C_{6}H_{12} O_{6}$ ) =-1275.0

enthalpy of combustion of oxygen(Δ $H_{f}$ of $O_{2}$ ) = zero

enthalpy of combustion of carbon dioxide(Δ $H_{f}$ of $CO_{2}$ ) = -393.5

enthalpy of combustion of water(Δ $H_{f}$ of $H_{2} O$ ) = -285.8

To solve :

standard enthalpy of combustion

We know;

Δ $H_{f}$ = ∈Δ $H_{f}$ (products) - ∈Δ $H_{f}$ (reactants)

$C_{6}H_{12} O_{6}$ (s) +6 $O_{2}$ (g) → 6 $CO_{2}$ (g)+ 6 $H_{2} O$ (l)

Δ $H_{f}$ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

Δ $H_{f}$ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

Δ $H_{f}$ = 6 (-393.5) + 6(-285.8) - 0 + 1275

Δ $H_{f}$ = -2361 - 1714 - 0 + 1275

Δ $H_{f}$ =-2800 kJ

Therefore the standard enthalpy of combustion is -2800 kJ

7 0

2 years ago

2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo

pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

Chemical reaction involved: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: $Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2$

To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:

$\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}$

$\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)$

$\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)$

Therefore, the Gibb's free energy change at 37° C (310.15 K): ΔG = (-2.45 kJ/mol)

4 0

2 years ago