Is 5O2 a Element or Compound?

How many neutrons are present in an atom of

Tamiku [17]

There’s going to be 8 neutrons presented

4 0

3 years ago

Which best describes nuclear fusion? O A nucleus spontaneously splits and absorbs energy O Two nuclei spontaneously combine and

KIM [24]

Answer: Nuclei combine to form a heavier nucleus, releasing energy. 

Explanation: e.g two deuterium nucleus (Hydrogen-2 isotopes) forms an He nucleus and energy is released.

4 0

2 years ago

Mg(s) + HCl (l) --> H2 (g) +MgCl2 (s) A student conducted an experiment with 2g of magnesium and 3 mL of 6M hydrochloric acid

alexira [117]

Answer: Do you have an image so I can solve it?

7 0

2 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

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

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

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, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 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 = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

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 = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

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

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

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

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

2 years ago

What elements are found in salt?

devlian [24]

Sodium is a member of the alkali metal family with potassium (K) and Lithium (LI) sodium's big claim to fame is that it's one or two elements in your table salt. when bonded to chlorine (CI) THE two elements make sodium chloride

8 0

2 years ago