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Anna [14]
3 years ago
14

Ben is pushing a small trolley 10 N to the north and but Gary is pushing the trolley 5 N to the south. What is the total applied

force on the trolley?
Chemistry
1 answer:
andrew-mc [135]3 years ago
5 0

Answer:

The total applied force on the trolley is 5 N toward the North.

Explanation:

In this problem, Ben is pushing the trolley to north and Gary is pushing it to south. So both the forces are acting 180° opposite to each other. As force is a vector quantity, the net force or total force acting on any object should be calculated by vector addition of number of forces along with their directions. So in this case, if we consider the force Ben is applying as F1 and the force Gary is applying as F2 on the trolley. Then the net or total force acting on the trolley will be F1-F2. This is because, F1 and F2 are acting opposite to each other in direction.Thus, Total force acting on the trolley = 10 N - 5 N.

So the total force acting on the trolley is 5 N and it is toward the north direction.

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kolezko [41]

Answer:

Cl2(g) (green/yellow mix) + 2KBr(s) (white) ---> 2KCl(s) (violet) + Br2(g) (reddish brown)

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Explanation:

Look at the oxidation state, when the number increase your element gets oxidated, when the number decrease, the elements it's getting reduced.

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2 years ago
Many power plants produce energy by burning carbon-based fuels, which also produces carbon dioxide. Carbon dioxide is a greenhou
RUDIKE [14]

Answer:

a) 2.541 mol/MJ;

b) 1.124 mol/MJ;

c) 0.4354 mol/MJ;

d) 0.1835 mol/MJ

Explanation:

The enthalpy of formation (ΔH°f) is the enthalpy of a reaction to form a compound by its constituents. For CO₂, ΔH°f = - 393.5 kJ/mol.

The enthalpy of a reaction is the sum of the enthalpy of the products (each one multiplied by the number of moles) less the sum of the enthalpy of the reactants (each one multiplied by the number of moles). The ΔH°f for simple substances (with one atom) is 0. The combustion is the reaction between the fuel and the oxygen.

a) The combution reaction is:

C(s) + O₂(g) → CO₂(g)

ΔH°rxn = -393.5 kJ/mol = -393.5x10⁻³ MJ/mol

Number of moles per MJ released: 1/|ΔH°rxn|

n = 1/(393.5x10⁻³) = 2.541 mol/MJ

b) The combustion reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

H₂O is in the liquid state because it's at 1 atm and 25ºC.

ΔH°f, H₂O(l) = -285.3 kJ/mol

ΔH°f, O₂(g) = 0

ΔH°f, CH₄(g) = -74.8 kJ/mol

ΔH°rxn = [2*(-285.3 ) + 1*(-393.5)] - [1*(-74.8)]

ΔH°rxn = -889.3 kJ/mol = -889.3x10⁻³ MJ/mol

n = 1/889.3x10⁻³ = 1.124 mol/MJ

c) C₃H₈(g) + 10O₂(g) → 3CO₂(g) + 4H₂O(l)

ΔH°f,C₃H₈(g) = -25.2 kJ/mol

ΔH°rxn = [4*(-285.3) + 3*(-393.5)] - [1*(-25.2)]

ΔH°rxn = -2,296.5 kJ/mol = -2.2965 MJ/mol

n = 1/2.2965 = 0.4354 mol/MJ

d) C₈H₁₈(l) + (25/2)O₂(g) → 8CO₂(g) + 9H₂O(l)

ΔH°f, C₈H₁₈(l) = -250.1 kJ/mol

ΔH°rxn = [9*(-283.5) + 8*(-393.5)] - [1*(-250.1)]

ΔH°rxn = -5,449.4 kJ/mol = -5.4494 MJ/mol

n = 1/5.4494 = 0.1835 mol/MJ

4 0
3 years ago
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Maksim231197 [3]

Answer:

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Explanation:

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allsm [11]

Answer:

The final balanced equation is :

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Explanation:

SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)

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First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)

Balance the charge by adding 2 electrons on product side:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-....[1]

Reduction :

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)

Balance O by adding water on required side:

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

Now, balance H by adding H^+ on the required side:

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At last balance the charge by adding electrons on the side where positive charge is more:

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Adding [1] and [2]:

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

4 0
3 years ago
1. What is the relation between distance and speed?
Inga [223]

Answer:

option no b is right answer

6 0
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