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3 years ago

- What is the mass of the following particles: o Protons o Neutrons o Electrons

Chemistry

1 answer:

anastassius [24]3 years ago

6 0

Answer:

the answer is:

Explanation:

Relative mass:

electron:0

proton: 1

neutron: 1

actual mass (g):

electron:9.11×10^-28

proton and neutron: 1.67 × 10^-24

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Anabolic reactions _______ bonds, whereas catabolic reactions __________ bonds. A. decrease; increase. B. break; make C. weaken;

nexus9112 [7]

Answer:

The corrext answer is E. make; break

Explanation:

In living organisms, the metabolism is either anabolic or catabolic where anabolic metabolism is energy consuming and catabolic metabolism is eneegy releasesing. It should however be noted that anabolic reaction builds or biosynthesize new mollecular structures while catabolic reaction breaks down complex structure bonds into simple structures

The braking down of bonds in catabolic reations realeses energy to sustain the anabolic rection process for the formation of new bonds

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3 years ago

How many grams of NaOH would be needed to make 0.50 L of a 1.25 molar NaOH solution

Likurg_2 [28]

The solution would be made by dissolving 4 grams NaOH in .5 Liters of H2O

3 0

3 years ago

What is oxidized and what is reduced in this reaction?

sineoko [7]

N⁻²2H4(l) + 2H2O⁻¹2(l) → N⁰2(g) + 4H2O⁻²(g)

N is oxidized and O is reduced

6 0

3 years ago

Calculate the value of the diffusion coefficient D (in m2/s) at 547°C for the diffusion of some species in a metal; assume that

svetlana [45]

Answer : The value of diffusion coefficient is, $2.97\times 10^{-16}m^2/s$

Explanation :

Formula used :

$D=D_o\times \exp \left(-\frac{Q_d}{RT}\right )$

where,

D = diffusion coefficient = ?

$D_o$ = $5.6\times 10^{-5}m^2/s$

$Q_d$ = $177kJ/mol$

R = gas constant = 8.314 J/mol.K

T = temperature = $547^oC=273+547=820K$

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

$D=5.6\times 10^{-5}m^2/s\times \exp \left(-\frac{177kJ/mol}{(8.314J/mol.K)\times (820K)}\right )$

$D=2.97\times 10^{-16}m^2/s$

Thus, the value of diffusion coefficient is, $2.97\times 10^{-16}m^2/s$

6 0

3 years ago

Alice and Bob are experimenting with two moles of neon, a monatomic gas, that starts out at conditions of standard temperature a

Archy [21]

Answer:

29273.178 joules have been added to the gas for the entire process.

Explanation:

The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

Isochoric (Constant volume)

$c_{v} = \frac{3}{2}\cdot R_{u}$ (1)

Isobaric (Constant pressure)

$c_{p} = \frac{5}{2}\cdot R_{u}$ (2)

Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) <em>Alice heats the gas at constant volume until its pressure is doubled</em>:

$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

$T_{2} = 2\times 273.15\,K$

$T_{2} = 546.3\,K$

(ii) <em>Bob further heats the gas at constant pressure until its volume is doubled</em>:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

$c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{v} = 12.471\,\frac{J}{mol\cdot K}$

$c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{p} = 20.785\,\frac{J}{mol\cdot K}$

$Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]$

$Q = 29273.178\,J$

29273.178 joules have been added to the gas for the entire process.

6 0

3 years ago