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joja [24]
3 years ago
13

Chemical reaction to form products this chemical equation represents one such reaction the coefficient for one of the reactants

or products is incorrect which part of the chemical equation is incorrect. 2c⁴h¹⁰ + 10o² -> 8co² +10h²o​
Chemistry
1 answer:
Dmitriy789 [7]3 years ago
4 0

Answer:

In a chemical equation, chemicals that react are the reactants, while chemicals that are produced are the products/by products. Both sides of the equation must be balanced.

\

Explanation

When writing a chemical equation, reactants reacts to produce products. For example in the equation for formation of water, hydrogen combines with oxygen as 2H₂ +O₂→2H₂O where the first part before the arrow represent the reactants and the next part after the arrow are the products. Reactants are on the left where as products are on the right.Coefficient 2, in this cases is used for balancing the equation

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Both hydrogen sulfide (H2S) and ammonia (NH3)
faust18 [17]

Answer: The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster.

Effusion rate is inversely proportional to molar mass.

NH3 will have a higher average molecule velocity, so it will diffuse faster and will reach the other side of the room more quickly.

Explanation: change up your response a bit

4 0
3 years ago
Read 2 more answers
Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.
zubka84 [21]

<u>Answer:</u> The value of \Delta G^o of the reaction is 28.38 kJ/mol

<u>Explanation:</u>

For the given chemical reaction:

SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)

  • The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

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

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]

We are given:

\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol

  • The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]

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

\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]

We are given:

\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}

where,

\Delta H^o_{rxn} = standard enthalpy change of the reaction =-67200 J/mol

\Delta S^o_{rxn} = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol

Hence, the value of \Delta G^o of the reaction is 28.38 kJ/mol

7 0
3 years ago
How much energy is used to heat 250 g of ice from –15°C to steam at 105°C? (Hint: 5 steps, all positive)
lianna [129]

Answer:

Explanation: Q1 = mc(ice) ΔT (ice warms)

Q2 = ms (ice melts)

Q3 = mc((water) ΔT (water warms)

Q4 = mr (water boils)

Q5 = mc(vapour)ΔT

7 0
3 years ago
A 25.0g sample of brass, which has a specific heat capacity of 0.375·J·g−1°C−1, is dropped into an insulated container containin
Angelina_Jolie [31]

Answer:

The equilibrium temperature of water is 25.6 °C

Explanation:

Step 1: Data given

Mass of the sample of brass = 25.0 grams

The specific heat capacity = 0.375 J/g°C

Mass of water = 250.0 grams

Temperature of water = 25.0 °C

The initial temperature of the brass is 96.7°C

Step 2: Calculate the equilibrium temperature

Heat lost = heat gained

Q(sample) = -Q(water)

Q = m*C* ΔT

m(sample)*c(sample)*ΔT(sample) = - m(water)*c(water)*ΔT(water)

⇒m(sample) = the mass of the sample of brass = 25.0 grams

⇒with c(sample) =The specific heat capacity = 0.375 J/g°C  

⇒with ΔT = the change of temperature = T2 - T1 =T2 - 96.7 °C

⇒with m(water) = the mass of the water = 250.0 grams

⇒with c(water) = the specific heat capacity = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = T2 - 25.0°C

25.0 * 0.375 * (T2 - 96.7) = - 250.0 * 4.184 J/g°C * (T2 - 25.0°C)

9.375T2 - 906.56 = -1046T2 + 26150

9.375T2 + 1046T2 = 26150 + 906.56

1055.375T2 = 27056.26

T2 = 25.6 °C

The equilibrium temperature of water is 25.6 °C

4 0
3 years ago
I RLLY NEED HELP. Use Image B (picture above) and calculate the density of a ring that has a mass of 32 grams. Read Page 11 "Mea
dimaraw [331]

Answer:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring = 8 g/mL

Explanation:

From the question given above, the following data were obtained:

Mass of ring = 32 g

Volume of water = 64 mL

Volume of water + ring = 68 mL

Density of ring =?

Next, we shall determine the volume of the ring. This can be obtained as follow:

Volume of water = 64 mL

Volume of water + ring = 68 mL

Volume of ring =?

Volume of ring= (Volume of water + ring) – (Volume of water)

Volume of ring = 68 – 64

Volume of ring = 4 mL

Finally, we shall determine the density of the ring. This can be obtained as follow:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring =?

Density = mass / volume

Density of ring = 32 / 4

Density of ring = 8 g/mL

8 0
2 years ago
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