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Tju [1.3M]
2 years ago
13

Find the force it would take to accelerate an 800-kg car at a rate of 5m/s2.

Chemistry
2 answers:
yanalaym [24]2 years ago
8 0

Answer:

4000 N

Explanation:

According to Newton's second law, a resultant force is equal to mass times acceleration. i.e

F_{net} = ma = 800 kg * 5 m/s^{2}   = 4000 N

So assuming that there is no other force that is acting on this car, a force of 4000 N would cause the acceleration mentioned in the question.

laila [671]2 years ago
3 0
Force= mass•acceleration

Force=800•5

Force=4000 N
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Which of the following possess the greatest concentration of hydroxide ions?
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Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

<u>(a) a solution of pH 3.0</u>

First we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

11=-\log [OH^-]

[OH^-]=1.0\times 10^{-11}M

Thus, the OH^- concentration is, 1.0\times 10^{-11}M

<u>(b) a solution of 0.10 M NH_3</u>

As we know that 1 mole of NH_3 is a weak base. So, in a solution it will not dissociates completely.

So, the OH^- concentration will be less than 0.10 M

<u>(c) a solution with a pOH of 12.</u>

We have to calculate the OH^- concentration.

pOH=-\log [OH^-]

12=-\log [OH^-]

[OH^-]=1.0\times 10^{-12}M

Thus, the OH^- concentration is, 1.0\times 10^{-12}M

<u>(d) a solution of 0.10 M NaOH</u>

As we know that NaOH is a strong base. So, it dissociates to give Na^+ ion and OH^- ion.

So, 0.10 M of NaOH in a solution dissociates to give 0.10 M of Na^+ ion and 0.10 M of OH^- ion.

Thus, the OH^- concentration is, 0.10 M

<u>(e) a 1\times 10^{-4}M solution of HNO_2</u>

As we know that 1 mole of HNO_2 in a solution dissociates to give 1 mole of H^+ ion and 1 mole of NO_2^- ion.

So, 1\times 10^{-4}M of HNO_2 in a solution dissociates to give 1\times 10^{-4}M of H^+ ion and 1\times 10^{-4}M of NO_2^- ion.

The concentration of H^+ ion is 1\times 10^{-4}M

First we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (1.0\times 10^{-4})

pH=4

Now we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

10=-\log [OH^-]

[OH^-]=1.0\times 10^{-10}M

Thus, the OH^- concentration is, 1.0\times 10^{-10}M

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3 years ago
Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241
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Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)     \Delta H_1=-241.8kJ

(2) X(s)+2Cl_2(g)\rightarrow XCl_4(s)    \Delta H_2=+461.9kJ

(3) \frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)    \Delta H_3=-92.3kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) H_2O(g)\rightarrow H_2O(l)    \Delta H_5=-44.0kJ

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) 2H_2O(g)\rightarrow 2H_2(g)+O_2(g)     \Delta H_1=2\times 241.8kJ=483.6kJ

(2) XCl_4(s)\rightarrow X(s)+2Cl_2(g)    \Delta H_2=-461.9kJ

(3) 2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)    \Delta H_3=4\times -92.3kJ=-369.2kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) 2H_2O(l)\rightarrow 2H_2O(g)    \Delta H_5=2\times 44.0kJ=88.0kJ

The expression for enthalpy of main reaction will be:

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5

\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)

\Delta H=-1048.6kJ

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

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