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2 years ago

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In Part A, you found the number of moles of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxyge

n. In Part B, you found the number of moles of product (1.40 mol P2O5 ) formed from the given amount of oxygen and excess phosphorus.

1 answer:

Shtirlitz [24]2 years ago

5 0

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18. What is the percent by mass sugar of a solution made by dissolving 12.45 grams of sugar in 100 grams of water?

Fantom [35]

The percent by mass sugar of a solution : 11.07%

<h3>Further explanation</h3>

Given

mass of sugar = 12.45 g

mass of water = 100 g

Required

The percent by mass

Solution

Mass of solution :

$\tt mass~sugar+mass~water=12.45+100=112.45~g$

Percent mass of sugar :

$\tt \%mass=\dfrac{mass~sugar}{mass~solution}\times 100\%\\\\\%mass=\dfrac{12.45}{112.45}\times 100\%=11.07\%$

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3 years ago

Write the balanced equation. Then calculate the volume of 0.65 M HCl required to completely neutralize 400.0 ml of 0.88 M KOH.

Misha Larkins [42]

Hello!

The balanced equation for the neutralization of KOH is the following:

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

To calculate the volume of HCl required, we can apply the following equation:

$moles HCl = moles KOH \\ \\ cHCl*vHCl=cKOH*vKOH \\ \\ vHCl= \frac{cKOH*vKOH}{cHCl}= \frac{400 mL*0,88M}{0,65M}= 541,54mL$

So, the required volume of HCl is 541,54 mL

Have a nice day!

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3 years ago

calculate δg o for each reaction using δg o f values: (a) h2(g) i2(s) → 2hi(g) 2.6 kj (b) mno2(s) 2co(g) → mn(s) 2co2(g) kj (c)

Reika [66]

(a)The change in Gibbs free energy for the reaction has been 2.6 kJ/mol.

(b) The change in Gibbs free energy for the reaction has been -49.3 kJ/mol.

(c) The change in Gibbs free energy for the reaction has been 91.38 kJ/mol.

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2 years ago

the smell spreads immediately after you burn agarbatti in this example what is your condition about intermolecular force

Sloan [31]

Intermolecular force for solids is high. Whereas low in gases. The smell of agarbatti spreads immediately because the molecules of air diffuses very fastly.

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3 years ago

Read 2 more answers

Nitric monoxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO₂), a dark brown gas. If 5.895 mol of NO is mixed with 2

Schach [20]

Answer:

Limiting reactant: O2

grams NO2 produced = 230.276 g NO2

grams of NO unused = 26.67 gNO

Explanation:

2NO + O2 --> 2NO2

Step 1: Determine the molar ratio NO:O2

molar ratio NO:O2 = 5.895: 2.503 = 2.35

stoichiometric molar ratio NO:O2 = 2:1

So, O2 is the limiting reactant.

Step2: Determine the grams of NO2:

?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2

Step 3: Determine the amount of excess reagent unreacted

moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted

moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted

mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted

8 0

3 years ago