Which property of a body cannot be changed if a force is applied to it?

How many moles of potassium chloride are in 28 grams of KCl?

stiv31 [10]

Answer:

0.3758moles

Explanation:

moles of kcl = mass of kcl/ molar mass of kcl = 28/74.5 = 0.3758moles

7 0

3 years ago

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N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol

docker41 [41]

Answer:

A

Explanation:

Recall that ΔH is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

$\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)$

Therefore, from the chemical equation, we have that:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}$

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}$

In conclusion, our answer is A.

5 0

2 years ago

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago

How does ionic radii increase

nalin [4]

<>"One such trend is closely linked to atomic radii -- ionic radii. Neutral atoms tend to increase in size down a group and decrease across a period. When a neutral atom gains or loses an electron, creating an anion or cation, the atom's radius increases or decreases, respectively."<>

3 0

3 years ago

1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

Rasek [7]

Answer:

Q1: 728.6 J.

Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:

The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

∴ The amount of heat absorbed by water = Q = m.c.ΔT = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

Q2: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.

a) First part: the energy change (q) of the surroundings (water):

The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

∴ The amount of heat absorbed by water = Q = m.c.ΔT = (25 g)(4.18 J/g°C)(6.4°C) = 668.8 J.

b) second part:

Q water = Q unknown metal.

Q unknown metal = - 668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

Q unknown metal = - 668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

- 668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (- 668.8)/(- 1914) = 0.3495 J/g°C.

3 0

3 years ago