What is the most common ion for cesium ? Cs+1 Cs Cs-1 Cs-2

The reaction A + 2B occurs in one step in the gas phase. In each blank below, write the exponent of the concentration in the FOR

strojnjashka [21]

Answer:

see below

Explanation:

for A + 2B => Products ...

Rate Law => Rate =k[A][B]ˣ

As shown in expression, A & B are included, C is not.

6 0

3 years ago

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According to the law of conservation of mass in a chemical reaction the total starting mass of all the reactants equals the tota

Ber [7]

I believe the correct answer true. According to the law of conservation of mass, in a chemical reaction the total starting mass of all the reactants equals the total final mass of all the products. This law states that mass cannot be created or be destroyed. So, the total mass that goes in a process should be equal to the mass that goes out the process. This is true for chemical reactions and physical processes. It is Antoine Lavoisier who described this and is a basic principle used in physics and in chemistry. Mass, unlike energy, cannot be transformed to any form so however the transformation happens the mass should be constant.

3 0

3 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0

4 years ago

How do you find time when you have been given the distance

MrRa [10]

Answer:To solve for time, divide the distance traveled by the rate. For example, if Cole drives his car 45 km per hour and travels a total of 225 km, then he traveled for 225/45 = 5 hours.

Explanation:

4 0

3 years ago

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Which statement correctly describes the Ka of a weak acid?

katovenus [111]

The answer is D) The Ka is small, and the equilibrium favors reactants.

6 0

3 years ago

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