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steposvetlana [31]
4 years ago
15

A scientist measures the standard enthalpy change for the following reaction to be -327.2 kJ : P4O10(s) 6 H2O(l)4H3PO4(aq) Based

on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of P4O10(s) is kJ/mol.
Chemistry
1 answer:
OleMash [197]4 years ago
5 0

Answer:

ΔH°f P4O10(s) = - 3115.795 KJ/mol

Explanation:

  • P4O10(s) + 6H2O(l) ↔ 4H3PO4(aq)
  • ΔH°rxn = ∑νiΔH°fi

∴ ΔH°rxn = - 327.2 KJ

∴ ΔH°f H2O(l) = - 285.84 KJ/mol

∴ ΔH°F H3PO4(aq) = - 1289.5088 KJ/mol

⇒ ΔH°rxn = (4)(- 1289.5088) - (6)(- 285.84) - ΔH°f P4O10(s) = - 327.2 KJ

⇒ ΔH°f P4O10(s) = - 5158.035 + 1715.04 + 327.2

⇒ ΔH°f P4O10(s) = - 3115.795 KJ/mol

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mrs_skeptik [129]

Answer:

Mass = 0.139 kg

Explanation:

There is a formula in terms of force, acceleration and mass which is:

Force = mass × acceleration

Put the values into the formula.

5 = m × 36

m = 5 ÷ 36

<u>m = 0.139 kg</u>

6 0
3 years ago
A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and
Dmitry_Shevchenko [17]

Answer:

1. Rate =k [NO]^{2}[Cl_{2}]

2. k= 0.42 \frac{L^{2}}{mol^{2}*s}

Explanation:

Rate =k [NO]^{m}[Cl_{2}]^{n}

Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2

Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1

Rate =k [NO]^{2}[Cl_{2}]^{1}

Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12}  = 0.42 \frac{L^{2}}{mol^{2}*s}

6 0
3 years ago
If 185 mg of acetaminophen were obtained from a tablet containing 350 mg of acetamino- phen, what would be the weight percentage
Alex73 [517]

Percentage recovery gives us an idea of the amount of pure substance recovered after the chemical reaction. Percentage recovery can be more than 100 % or less than 100 %. Usually, in any experiment performed the weight percentage recovery will be less than 100. Percent recovery values greater than 100 show that the recovered compound is contaminated.

Amount of acetaminophen initially taken = 350 mg

Amount of acetaminophen obtained after recovery =185 mg

Weight percentage recovery =\frac{mass recovered}{mass originally taken}*100

                                                = \frac{185 mg}{350 mg}*100

                                               = 52.9%

8 0
3 years ago
Describe ways a lack of water could affect cell functions?
ladessa [460]

The way a lack of water could affect cell functions is that cell membranes would lack structure.

Osmosis causes water to flow from the cell into the surrounding liquid. This causes the protoplasm all the material inside the cell, to shrink from the cell wall. Excessive water loss leading to cell wall breakdown can lead to cell death. Water is an important part of the protoplasm of living cells as it is directly involved in myriad biochemical reactions such as photosynthesis and respiration.

Without them, cells cannot move waste and byproducts absorb nutrients, carry out intracellular transport functions or send signals. Regulates body temperature. Moisten eye nose and mouth tissues. Protects body organs and tissues. Carry nutrients and oxygen to cells. Plasmolysis is a situation in which cells lose water in hypertonic solutions. the plant cells lose water causing them to shrink or shrink removing the cell cytoplasm from the cell wall.

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4 0
1 year ago
What is the mass in grams of 5.00moles of CH4?
anastassius [24]

Answer:

1. 80g

2. 1.188mole

Explanation:

1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Number of mole of CH4 from the question = 5 moles

Mass of CH4 =?

Mass = number of mole x molar Mass

Mass of CH4 = 5 x 16

Mass of CH4 = 80g

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Molar Mass of O2 = 16x2 = 32g/mol

Number of mole O2 =?

Number of mole = Mass /Molar Mass

Number of mole of O2 = 38/32

Number of mole of O2 = 1.188mole

6 0
3 years ago
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