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ANEK [815]
3 years ago
15

My teacher laid this much out for us but I don’t know how to get the products in each one.

Chemistry
1 answer:
irakobra [83]3 years ago
3 0

Answer:dang

Explanation:

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Ammonia (NH3) boils at -33∘C; at this temperature it has a density of 0.81 g/cm3. The enthalpy of formation of NH3(g) is -46.2 k
Minchanka [31]

Answer:

-87098.82kJ

Explanation:

Hello,

This process could be attained into two steps due to the specified conditions at assuming that it is at -33°C at the beginning:

ΔH=ΔvapH+ΔrH

Thus:

ΔvapH=5L*0.81\frac{kg}{L}*\frac{1000g}{1kg}*\frac{1molNH_3}{17gNH_3}*23.2\frac{kJ}{molNH_3}=5527.06kJ

Once the ammonia is gaseous, the burning chemical reaction is:

NH_3+\frac{3}{4} O_2-->\frac{1}{2}N_2+\frac{3}{2}H_2O

So the enthalpy of reaction for the given amount of ammonia is:

ΔrH=5L*0.81\frac{kg}{L}*\frac{1000g}{1kg}*\frac{1molNH_3}{17gNH_3}*(\frac{3}{2}*-290\frac{kJ}{mol}-(-46.2\frac{kJ}{mol})  ) =-92625.88kJ

Finally, the total enthalpy is:

ΔH=5527.06kJ-92625.88kJ=-87098.82kJ

Best regards.

7 0
3 years ago
Write a balanced half-reaction for the reduction of permanganate ion (Mno) to manganese ion Mn?) in acidic aqueous solution. Be
Luden [163]

Explanation:

Permanganate is general name for the chemical compound which containes manganate (VII) ion. Permanganate(VII) ion is strong oxidizing agent as  manganese is in +7 oxidation state and can be easily reduced and oxidize others.

The balanced half reaction for reduction of the permanganate ion, MnO_4^- to manganese ion, Mn^{2+} is shown below:

MnO_4^-_{(aq)}+8H^+_{(aq)}+5e^-\rightarrow Mn^{2+}_{(aq)}+4H_2O_{(l)}

6 0
3 years ago
What would happen to the boiling point of water at 8,000 m above sea level, where air pressure is lower?
mario62 [17]
it will get warm or spread out into the water or
5 0
3 years ago
Suppose there was a release of 1 mole of Alpha emission particle and 1 mole of Beta emission particles and both particles are ac
riadik2000 [5.3K]

Explanation:

translate the following word equation in the form of balance come alive with your number is aluminium + hydrochloric acid and Aluminium chloride + nitrogen

7 0
3 years ago
2.643 grams of potassium butanoate (KCH3(CH2)2CO2 ) is fully dissolved in 50.00 mL of water, which is carefully transferred to a
lara [203]

Answer:

The pH of the solution is 4.69

Explanation:

Given that,

Mass of potassium = 2.643 grams

Weight of water = 50.00 mL

Weight of HCl=100.00 ml

Mole = 0.120 M

We know that,

KCH_{3}(CH_{2})_{2}CO_{2} is a basic salt.

Let's write it as KY.

The acid HCH_{3}(CH_{2}CO_{2}) would become HY.

We need to calculate the moles of KY

Using formula of moles

moles\ of\ KY=\dfrac{m}{M}\times1000

moles\ of\ KY=\dfrac{2.643}{126}\times1000

moles\ of\ KY=20.97\ m\ mole

The reaction is

KY+HCl\Rightarrow HY+ KCl

The number of moles of KY is 20.98 m

initial moles = 20.98

Final moles m=20.98-0.120\times100= 8.98

We need to calculate the value of pKa(HY)

Using formula for pKa(HY)

pKa_{HY}=-log Ka

pKa_{HY}=-log(1.5\times10^{-5})

pKa_{HY}=4.82

We need to calculate the pH of the solution

Using formula of pH

pH=pKa+log(\dfrac{[KY]}{[KH]})

Put the value into the formula

pH=4.82+log(\dfrac{8.98}{12})

pH=4,69

Hence, The pH of the solution is 4.69

7 0
3 years ago
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