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bekas [8.4K]
3 years ago
14

Two or more velocities added by......​

Chemistry
1 answer:
alexdok [17]3 years ago
5 0

speed and direction in which an object is moving. both speed and direction of motion. is a vector. two or more velocities add by velocity addition.

You might be interested in
How many hydrogen atoms are attached to each carbon adjacent to a double bond?
nekit [7.7K]
Carbon can make four bonds. Since it is doubly bonded carbon, it means it can make two more bonds. So, two more hydrogen atoms could help in bond formation.

H_{2}C=CH_{2}

Two H atoms are attached to each carbon.
6 0
3 years ago
rite an equation for the formation one mole of CaCO3(s) from its elements in their standard states. Write any reference to carbo
maks197457 [2]

Answer: The chemical equation for the formation one mole of CaCO3(s) from its elements in their standard states is Ca(s)+C(s)+\frac{3}{2}O_2(g)\rightarrow CaCO_3(s)

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

The chemical equation for the formation one mole of CaCO3(s) from its elements in their standard states is:

Ca(s)+C(s)+\frac{3}{2}O_2(g)\rightarrow CaCO_3(s)

where (s) represents solid state and (g) represents gaseous state.

6 0
3 years ago
For the following systems at equilibrium C: CaCO3(s) ⇌ CaO(s)+CO2(g) ΔH=+178 kJ/mol D: PCl3(g)+Cl2(g) ⇌ PCl5(g) ΔH=−88 kJ/mol cl
Rama09 [41]

Explanation:

C: CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)ΔH=+178 kJ/mol

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

A+\text{heat}\rightleftharpoons B

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.  This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Treat heat as a reactant and on increasing a reactant at equilibrium, shifts the reaction in the forward direction.

Increase temperature →  increase in heat → forward direction

Decrease temperature →  decease in heat → backward direction

System C - Increase temperature : Reaction will move forward

System C - Decrease temperature : Reaction will move backward

D: PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g) ΔH=−88 kJ/mol

The total enthalpy of the reaction comes out to be negative .

The temperature of the surrounding will increase.

For an exothermic reaction, heat is released during a chemical reaction and is written on the product side.

A\rightleftharpoons B+\text{ heat}

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.  This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Treat heat as a product and on increasing a product at equilibrium, shifts the reaction in the backward direction.

Increase temperature →  increase in heat → backward direction

Decrease temperature →  decease in heat → forward direction

System D - Increase temperature : Reaction will move backward

System D - Decrease temperature : Reaction will move forward

7 0
3 years ago
Find the [H+] in an acetic acid solution that has a pH of 5.12
Eva8 [605]
<span>Data:
pH = 5.2
[H+] = ?

Knowing that: (</span><span>Equation to find the pH of a solution)</span>
pH = -log[H+]
<span>
Solving:
</span>pH = -log[H+]
5.2 = - log [H+]
Knowing that the exponential is the opposite operation of the logarithm, then we have:
[H+] = 10^{-5.2}
\boxed{\boxed{[H+] = 6.30*10^{-6}}}\end{array}}\qquad\quad\checkmark
3 0
3 years ago
The decomposition of N_2O_5(g) following 1st order kinetics. N_2O_5(g) to N_2O_4(g) + ½ O_2(g) If 2.56 mg of N_2O_5 is initially
Crank

Answer: The rate constant is 0.334s^{-1}

Explanation ;

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = age of sample  = 4.26 min

a =  initial amount of the reactant  = 2.56 mg

a - x = amount left after decay process  = 2.50 mg

Now put all the given values in above equation to calculate the rate constant ,we get

4.26=\frac{2.303}{k}\log\frac{2.56}{2.50}

k=\frac{2.303}{4.26}\log\frac{2.56}{2.50}

k=5.57\times 10^{-3}min^{-1}=5.57\times 10^{-3}\times 60s^{-1}=0.334s^{-1}

Thus rate constant is [tex]0.334s^{-1}

4 0
3 years ago
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