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vagabundo [1.1K]
3 years ago
9

20 points

Chemistry
1 answer:
Rom4ik [11]3 years ago
5 0

the answer is aaaaaaaaaaaaaaaaaaaaaaa

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Please help question in the picture
bogdanovich [222]

Goto ----> prize.gg to claim your robux PRIZE!!!!

6 0
3 years ago
Read 2 more answers
9. How many grams are in 2.0 x 105 molecules of carbon monoxide?
timama [110]

Answer: 9.3 x 10^ 18 g CO

Explanation:

Start by knowing that carbon monoxide is the compound CO. To convert molecules to grams, you first need to convert molecules to moles. This can be done using the conversion factor for Avogadro's Number:

(2.0 x 10^5 molecules CO) x 1 mol CO / 6.02 x 10^23 molecules CO

This cancels molecules CO.

Then, you can convert moles to grams, which is your desired quantity. You can find the number of grams for CO by looking at the periodic table and adding together their masses. C = 12 g and O = 16 g. Total of 28 g CO:

(1 mol CO) x 28 g CO / 1 mol CO

This cancels mol CO, which leaves grams CO.

5 0
3 years ago
Increasing the temperature, as noted in the graph, increases the rate of this chemical reaction. What is the relationship betwee
Soloha48 [4]

Answer: (C) Although the average kinetic energy of the colliding substances increases, this has no influence on activation energy.

Explanation:

After increasing the temperature of the reaction , the rate of the chemical reaction increases due to increase in the average kinetic energy of the particles. At increased temperature high proportions of particles can react making the reaction faster.

8 0
3 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0
3 years ago
At 73.0 ∘c , what is the maximum value of the reaction quotient, q, needed to produce a non-negative e value for the reaction so
damaskus [11]
Here we will use the general formula of Nernst equation:

Ecell = E°Cell - [(RT/nF)] *㏑Q

when E cell is cell potential at non - standard state conditions

E°Cell is standard state cell potential = - 0.87 V

and R is a constant = 8.314 J/mol K

and T is the temperature in Kelvin = 73 + 273 = 346 K

and F is Faraday's constant = 96485 C/mole

and n is the number of moles of electron transferred in the reaction=2  

and Q is the reaction quotient for the reaction 
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔  Br2(aq) + SO2(g) +2H2O(l)

so by substitution :

0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q      → solve for Q 


∴ Q = 4.5 x 10^-26 
6 0
3 years ago
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