A certain isotope x2+ contains 48 electrons and 75 neutrons. what is the mass number for this element?

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alisha [2963]

Answer:

During this process, energy is released as a bond is formed.

Explanation:

The elements present on the left side of the reaction equation are known as reactants, while those on the right are called the products.

In this case, N(g) and N(g) are the reactants, and N2(g) along with energy are the products.

Post-reaction, N(g) has vanished and has been replaced by N2(g). It confirms that a bond exists between the two N(g).

Moreover, energy has been released as part of the products (indicated on the right side)

The accurate statement is ''A bond is formed and energy is released''

5 0

3 months ago

Anarel [2989]

The balanced chemical equation for the neutralization of HCl with $NaHCO_{3}$ is:

$HCl (aq) + NaHCO_{3}(aq)--> NaCl (aq) + H_{2}O(l)+CO_{2} (g)$

Given weight of $NaHCO_{3}$ = 5g

Moles of $NaHCO_{3}$ = $5 g *\frac{1 mol}{84 g} = 0.05952 mol NaHCO_{3}$

Volume of HCl solution = $50 cm^{3} * \frac{1 mL}{1cm^{3}} = 50 mL$

Assuming the density of the solution is 1.0 g/mL

Mass of HCl solution = 50 g

Overall mass of the solution = 50 g + 5 g = 55 g

To find the heat of neutralization, we calculate:

Q = m C ΔT

where m equals the mass of the solution = 55 g

C represents the specific heat capacity of the solution = 4.184 $\frac{J}{g. ^{0}C}$

ΔT signifies the temperature change = 6.8 K = (6.8 - 273) C = -266.2 $^{0}C$

$Q = 55 g * 4.184 \frac{J}{g. K}(6.8K) = 1565 J$

The enthalpy of neutralization per mole of $NaHCO_{3}$

= $\frac{1565J}{0.05952 mol} = 26294 \frac{J}{mol}*\frac{1 kJ}{1000J} =26.294kJ/mol$

3 0

4 months ago