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SashulF [63]
3 years ago
8

What is the name of this unstable isotope from number 9?

Chemistry
2 answers:
raketka [301]3 years ago
6 0

Is there a picture of the isotope or?- becaue I can’t help if I don’t have a visual.

den301095 [7]3 years ago
3 0
Is there anything else or even a other picture
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How many atoms of hydrogen combine with one carbon atom to form methane
bonufazy [111]

Answer:four hydrogen atoms

Explanation:

4 0
3 years ago
Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta
AysviL [449]

<u>Answer:</u> The \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]

We are given:

\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ

Putting values in above equation, we get:

-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ

Hence, the \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

8 0
3 years ago
HELP Me Please I will give 20 points
lisabon 2012 [21]

Answer:

1c

2a

3b

Explanation:

5 0
3 years ago
Read 2 more answers
Is it necessary to prepare sodium extract to detect nitrogen in urea ?​
Vilka [71]

Answer:

I think No

Explanation:

7 0
2 years ago
F
Katarina [22]

The statement is true in this situation is C. The size of Ffric is the same as the size of Fapp:

From the diagram, since the body is in equilibrium, the sum of vertical forces equals zero. Also, the sum of horizontal forces equal zero.

So, ∑Fx = 0 and ∑Fy = 0

Since Fapp acts in the negative x - direction and Ffric acts in the positive x - direction,

∑Fx = -Fapp + Ffric = 0

-Fapp + Ffric = 0

Fapp = Ffric

Also, since Fgrav acts in the negative y - direction and Fnorm acts in the positive y - direction,

∑Fy = Fnorm + (-Fgrav) = 0

Fnorm - Fgrav = 0

Fnorm = Fgrav

So, we see that the size of Fapp <u>equals</u> size of Ffric and the size of Fnorm <u>equals</u> the size of Fgrav.

So, the correct option is C

The statement which is true in this situation is C. The size of Ffric is the same as the size of Fapp.

Learn more about equilibrium of forces here:

brainly.com/question/12980489

5 0
2 years ago
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