How many sig figs 18.01528

How many calories is 52 J of energy?

just olya [345]

Answer:

1J = 0.238846cal

52J = 12.42cal

5 0

3 years ago

What is the percent of hydrogen by mass in CH4O

WITCHER [35]

Answer:

My guess would be 12.5828

4 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

Write the balanced oxidation-reduction reaction equation for the oxidation of benzoin by ammonium nitrate. 2

Margarita [4]

Benzoin is an organic compound with the formula PhCH(OH)C(O)Ph. It is a hydroxy ketone attached to two phenyl groups.

To answer your question, the balanced oxidation-reduction reaction equation for the oxidation of benzoin by ammonium nitrate is:

2Ph-C(OH)-C(O)-Ph+NH4NO3 --> 2Ph-C(O)-C(O)-Ph + N2 + 3H2O.

I hope this helps and if you have any further questions, please don’t hesitate to ask again.

7 0

3 years ago

If temperature is increased , the number of collisions per second

AfilCa [17]

Answer: when the temperature is increased, the number of collisions per second increases.

Explanation:

the rate of collisions and the temperature is directly proportional. If the energy of the gas particles is boosted by using the temperature, the chances of the particles bumping into each other due to the high energy increases, thus increasing the number of collisions. This also increases the rate of reaction. Thus when temperature is increased the number of collisions also increases.

4 0

3 years ago