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3 years ago

How do isotopes of a givin element similar

2 answers:

6 0

Different isotopes of an element have the same number of protons in the nucleus, giving them the same atomic number, but a different number of neutrons giving each elemental isotope a different atomic weight.

BlackZzzverrR [31]3 years ago

5 0

Answer:

All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons.

Explanation:

Hope this Helped! :D

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A molecule of an organic compound contains at least one atom of

Mila [183]

A molecule of an organic compound contains at least one atom of "(1) carbon". This is referred to as organic chemistry, since all living matter is carbon-based.

5 0

3 years ago

Read 2 more answers

Consider the reaction H2(g) + I2(g) Double headed arrow. HI(g) with an equilibrium constant of 46.3 and a reaction quotient of 5

Alika [10]

Answer:

The reaction shifts to the left.

Explanation:

Equilibrium constant (K) = 46.3

Reaction Quotient (Q) = 525

The relationship between Q and K with their implications are given as;

K = Q (No net reaction)

K > Q (Reaction shifts to the right)

K < Q (Reaction shifts to the left)

Since in this question, Q (525) > K (46.3)

The reaction shifts to the left.

5 0

2 years ago

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

8 0

3 years ago

Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide

Contact [7]

Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

According to question

2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

$'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\$

Since 'NO' is the limiting reagent according to this ratio.

According to equation

2 moles NO reacts to form 2 moles NO₂

So, 0.1855 moles NO give = 0.1855 moles of NO₂

Mass of 1 mole NO₂ = 46 g/mole

Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g

5 0

3 years ago

On a recent trip to the Fort Worth Zoo, Trevor's parents bought him a helium-filled balloon. At the beginning of the day, the sk

timurjin [86]

Answer:

1. The new volume is 14L

2. The volume decreased

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 14.2 L

Initial pressure (P1) = 102.5 kPa

Initial temperature (T1) = 33°C

Final pressure (P2) = 100.9 kPa

Final temperature (T2) = 24°C.

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

K = °C + 273

T1 = 33°C = 33°C + 273 = 306K

T2 = 24°C = 24°C + 273 = 297K

Step 3:

Determination of the new volume.

Applying the general gas equation P1V1/T1 = P2V2/T2, the new volume can be obtain as follow:

P1V1/T1 = P2V2/T2

102.5 x 14.2/306 = 100.9 x V2/297

Cross multiply to express in linear form as shown below:

306 x 100.9 x V2 = 102.5 x14.2 x297

Divide both side by 306 x 100.9

V2 = (102.5x14.2x297)/(306 x 100.9)

V2 = 14 L

The new volume is 14 L.

Step 4:

Determination of the change in volume. This is illustrated below

Final volume (V2) = 14 L

Initial volume (V1) = 14.2 L

Change in volume (ΔV) =?

Change in volume (ΔV) = Final volume (V2) - Initial volume (V1)

ΔV = V2 - V1

ΔV = 14 - 14.2

ΔV = - 0.2L

Since the change in volume is negative, it means there is a decrease in the volume.

7 0

3 years ago

Read 2 more answers