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3 years ago

I need help. Like, I really dont know!!!

Chemistry

2 answers:

kap26 [50]3 years ago

7 0

Answer:

GIve me insta

Explanation:

zavuch27 [327]3 years ago

3 0

Answer: Ooo you be looking like a snack :D

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Calculate the average bond order for a p−o bond (such as the one shown in blue) in a phosphate ion. express your answer numerica

PtichkaEL [24]

<span>Answer: 1/4 is the average bond order for a pâ’o bond (such as the one shown in blue) in a phosphate ion.</span>

5 0

3 years ago

Diagonal lines in the graph represent the temperature of the substance is ___.

labwork [276]

Gdnndjfndmnxndndndjdjdjxncncncnnc

5 0

3 years ago

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)

nikdorinn [45]

Answer : The volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.

Explanation :

The balanced chemical reaction will be:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

First we have to calculate the moles of Al.

$\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}$

Molar mass of Al = 27 g/mole

$\text{Moles of }Al=\frac{55g}{27g/mol}=2mole$

Now we have to calculate the moles of $O_2$ .

From the reaction we conclude that,

As, 4 mole of Al react with 3 moles of $O_2$

So, 2 mole of Al react with $\frac{3}{4}\times 2=0.75\times 2$ moles of $O_2$

Now we have to calculate the volume of $O_2$ consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of $O_2$ occupies 22.4 liter volume of $O_2$ gas

So, $0.75\times 2$ mole of $O_2$ occupies $0.75\times 2\times 22.4$ liter volume of $O_2$ gas

Therefore, the volume of $O_2$ consumed are, $0.75\times 2\times 22.4$ L.

3 0

3 years ago

A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. the total heat capacity of the calorimeter plus water w

Sladkaya [172]

Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.

W are given a chemical reaction:

$Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)$

$c=5760J/^oC$

$\Delta T=0.570^oC$

To calculate the enthalpy change, we use the formula:

$\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J$

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = $\frac{3283.2J}{0.1326g}=24760.181J/g$

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

$\Delta H=24760.181J/g\times 24g/mol\\\\\Delta H=594244.3J/mol\\\\\Delta H=594.244kJ/mol$

4 0

3 years ago

If 8.00 g NH4NO3 is dissolved in 1000 g of water, the water decreases in temperature from 21.00 degrees Celsius to 20.39 degrees

telo118 [61]

Answer:

25.7 kJ/mol

Explanation:

There are two heats involved.

heat of solution of NH₄NO₃ + heat from water = 0

q₁ + q₂ = 0

n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃

∴ n = 0.100 mol NH₄NO₃

q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln

m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g

q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J

q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0

ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol

7 0

3 years ago