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3 years ago

if 980 KJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be

Chemistry

1 answer:

uranmaximum [27]3 years ago

6 0

Answer:

The final temperature of the water will be 328.81 K .

Explanation:

Using the equation, q = mcΔT

here, q = energy

m= mass

c= specific heat capacity

ΔT= change in temperature

Mass of water = 1kg (1000 g ) per liter

∴ 6.2 Liter of water = 6200 g

c of water ≈ 4.18 J /g/K

Now,

980000 = 6200*4.18*ΔT

ΔT = 37.81 K

∴ final temperature of the water = 291 + 37.81 = 328.81 K

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If 20.0g of water at 56.2℃ absorbs 195 J of heat, what is the final temperature? *

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Answer

Explanation:

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3 years ago

The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane wi

Setler79 [48]

Answer:

Explanation:

mass fraction N₂ : He : CH₄ : C₂H₆ : : 15 : 5 : 60 : 20

mole fraction N₂ : He : CH₄ : C₂H₆ : : 15/28 : 5/4 : 60/16 : 20/30

mole fraction N₂ : He : CH₄ : C₂H₆ : : .5357 : 1.25 : 3.75 : .67

Total mole fractions = .5357 + 1.25 + 3.75 + 0.67 = 6.2057

mole fraction of N₂ = .5357 / 6.2057 = .0877

mole fraction of He = 1.25 / 6.2057 = .20

mole fraction of CH₄ = 3.75 / 6.2057 = .6043

mole fraction of C₂H₆ = .67 / 6.2057 = .108

Partial pressure = total pressure x mole fraction

Partial pressure of N₂ = 1200 kPa x .0877 = 105.24 kPa

Partial pressure of He = 1200 kPa x .20 = 240 kPa

Partial pressure of CH₄ = 1200 kPa x .6043 = 725.16 kPa

Partial pressure of C₂H₆ = 1200 kPa x .108 = 129.6 kPa

6 0

3 years ago

How many moles are present in

olasank [31]

Answer: (a) There are 0.428 moles present in 12 g of $N_{2}$ molecule.

(b) There are 2 moles present in $12.044 \times 10^{23}$ particles of oxygen.

Explanation:

(a). The mass of nitrogen molecule is given as 12 g.

As the molar mass of $N_{2}$ is 28 g/mol so its number of moles are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{12 g}{28 g/mol}\\= 0.428 mol$

So, there are 0.428 moles present in 12 g of $N_{2}$ molecule.

(b). According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

Therefore, moles present in $12.044 \times 10^{23}$ particles are calculated as follows.

$Moles = \frac{12.044 \times 10^{23}}{6.022 \times 10^{23}}\\= 2 mol$

So, there are 2 moles present in $12.044 \times 10^{23}$ particles of oxygen.

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3 years ago

Jeremy had a set of 500 toy blocks that could be connected by snapping them together. At first, he used all the blocks to build

valentina_108 [34]

The masses were the same

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3 years ago

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Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

3 years ago