1. I think it is true?
2. Low melting points
3. True
4. Atomic number, I think it’s periods?
5. Groups?

Sorry, I might not get all of them right :(

Hope this helps you in any way!!

6 0

3 years ago

What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at −30.0 °C?

Wewaii [24]

Answer:

The total amount of heat released is 68.7 kJ

Explanation:

Given that:

mass of water = 94.0 g

moles of water = 94 / 18.02 = 5.216

80⁰C ------> 0⁰C --------> -30⁰C

Q1 = m Cp dT

= 94 x 4.184 x (0 - 80)

= -31463.68 J

= -31.43 kJ

Q2 = 6.01 x 10^3 x 5.216

= - 31348.16 J

= -31.35 kJ

Q3 = - 94 x 2.09 x 30

= - 5893.8 J

= -5.894 kJ

Total heat = Q1 + Q2 + Q3 = -31.43 kJ + (-31.35 kJ ) + (-5.894 kJ ) = -68.7 kJ

Total heat released = -68.7 kJ

Note that the "negative sign" simply indicates heat released, therefore no need to put it in the answer.

6 0

3 years ago

or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und

beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

$C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)$

$\Delta H^{o}_{rxn}= -5104.1kJ/mol$

The expression for the entropy change for the reaction is as follows.

$\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]$

$\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol$

$\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol$

$\Delta H^{o}_{f}(O_{2})= 0kJ/mol$

Substitute the all values in the entropy change expression.

$-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]$

$-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})$

$\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol$

$=-220.1kJ/mol$

Therefore, The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

4 0

3 years ago

The balanced equation shows how sodium chloride reacts with silver nitrate to form sodium nitrate and silver chloride. NaCl + Ag

Verdich [7]

<span>NaCl First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound Sodium = 22.989769 Chlorine = 35.453 Silver = 107.8682 Nitrogen = 14.0067 Oxygen = 15.999 Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results For NaCl 22.989769 + 35.453 = 58.44277 For AgNO3 107.8682 + 14.0067 + 3 * 15.999 = 169.8719 Now calculate how many moles of each substance by dividing the total mass by the molar mass For NaCl 4.00 g / 58.44277 g/mol = 0.068443 mol For AgNO3 10.00 g / 169.8719 g/mol = 0.058868 Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>

8 0

3 years ago