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vichka [17]
3 years ago
7

In which scenario is a chemical change occurring?

Chemistry
2 answers:
fenix001 [56]3 years ago
4 0

the answer is A because had this question before

Dimas [21]3 years ago
4 0
Answer is C because each of the above cause a change in state, hence it’s physical, the molecules remain same.
But corrosion is a chemical change, cause there is an addition of water and air to the regular iron molecules
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What is the molarity of 4.0 moles NaCI dissolved in 2.0 L of solution
Hunter-Best [27]

Answer:

What is the molarity of a solution containing 5.00 moles of kcl in 2.00L of solution? Molarity= moles of solute/volume of solution in litre , so the problem looks like this : 7/. 569 , which is equivalent to 12.302 M .

3 0
3 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}

(c) W_{isothermal}=5.743\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}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

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
If fluorine 20 undergoes beta decay , what will it become ?
e-lub [12.9K]
Fluorine 20 (F - Atomic number 9 and atomic mass 20). Firstly we need to know what is beta decay. Beta decay occurs when one neutron changes into a proton and an electron therefore the atomic mass will remain the same as even though we loose a neutron it is replaced by a proton, the atomic number is always raised by 1 when one beta decay occurs. The produced electron is shot out of the nucleus at an incredible speed. This speedy electron we call a beta particle.

Ok now the reaction.

 20       20        0
F  -> Ne    +     e
 9         10       -1

Remember the atomic number determines the nature of the element ( i.e what elemnt it is).
Hope this helps :).
7 0
3 years ago
HELP PLZ!!!!!
Molodets [167]

The answer would be c as the cart is not in motion therefor ruling out kinetic and it is completely at rest making all of it energy potential

6 0
3 years ago
The chemical equation below shows the amount of the reactants used and the amount of only one of the products formed. How much w
Karolina [17]

Answer:

CH4 + 2O2 → CO2 + 2H2O

Explanation:

This is all i could come up with im sorry.

6 0
2 years ago
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