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igor_vitrenko [27]
2 years ago
7

This means that the steel bar lost 14900 J of thermal energy. What is the change in temperature of the steel bar? Recall that th

e steel decreases in temperature, and use the value of 0.49 J/g•°C as the specific heat capacity of steel. The video shows the mass of the steel to be 40.7 g.
Chemistry
2 answers:
Basile [38]2 years ago
7 0

Answer:

\Delta T=-747.13^0C

Explanation:

Hello,

In this case, the heat is related with the change in the temperature via the heat capacity and the involved mass as shown below:

Q=mCp\Delta T

Now, as it is stated that there is an amount of lost heat, such value turn out negative implying that the final temperature is lower than the initial one, in such a way, one computes \Delta T as a measure of the change in the temperature as follows:

\Delta T=\frac{Q}{mCp} =\frac{-14900J}{0.49J/(g^0C)*40.7g}\\ \Delta T=-747.13^0C

Best regards.

stiv31 [10]2 years ago
5 0

Answer:

ΔT=-747,13°C

Explanation:

Sensible heat is<em> the amount of thermal energy that is required to change the temperature of an object</em>, the equation for calculating the heat change is  given by:

Q=msΔT

where:

  • Q, heat that has been absorbed or realeased by the substance [J]
  • m, mass of the substance [g]
  • s, specific heat capacity [J/g°C] (
  • ΔT, changes in the substance temperature [°C]

To solve the problem, we clear ΔT of the equation and then replace our data:

Q=msΔT

ΔT=Q/ms

ΔT=\frac{-14900 J}{40,7g*0,49\frac{J}{gC} }=-747,13°C

<em>(Note that Q=-14900 J because there is a </em><u><em>LOST</em></u><em> of thermal energy)</em>

Thus, the change in temperature of the steel bar is -747,13°C, meaning that the temperature of the bar decreases.

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How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?
lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

              N₂   +    3 H₂  ------—> 2 NH₃

             1 mol      3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

                     N₂             +           3 H₂        ------—>      2 NH₃

             1 mol (28 g/mol)     3 mol(2g/mol)

                    28 g                        6 g

28 grams of N₂ react with 6 g of H₂  

So

if 28 grams of N₂ produces 6 g of H₂  so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

                 28 g of N₂ ≅ 6 g of H₂

                 X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

                X g of N₂ = 28 g x 14.2 g / 6 g

                X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

                     N₂             +           3 H₂        ------—>      2 NH₃

                                                   6 g                            2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

                 6 g of H₂ ≅ 2 mol of NH₃

                14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

               X mol of NH₃= 14.2 g x 2 mol / 6 g

                X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

        mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of  NH₃

Molar mass of  NH₃ = 14 + 3(1)

Molar mass of  NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

        mass in grams = 4.73 mole x 17 g/mol

        mass in grams =  80.41 g

mass of NH₃=  80.41 g

3 0
2 years ago
his is the chemical formula for chromium(III) nitrate: . Calculate the mass percent of oxygen in chromium(III) nitrate. Round yo
Alinara [238K]

Answer:

\%\ Composition\ of\ iron=69.92\ \%

Explanation:

Percent composition is percentage by the mass of element present in the compound.

The formula for chromium(III) nitrate is Cr(NO_3)_3

Molar mass of chromium(III) nitrate = 238.011 g/mol

1 mole of chromium(III) nitrate contains 9 moles of oxygen

Molar mass of oxygen = 16 g/mol

So, Mass= Molar mass*Moles = 16*9 g = 144 g

\%\ Composition\ of\ iron=\frac{Mass_{iron}}{Total\ mass}\times 100

\%\ Composition\ of\ iron=\frac{144}{238.011}\times 100

\%\ Composition\ of\ iron=69.92\ \%

5 0
2 years ago
Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3
ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)×\frac{1molZn}{65,38g}×\frac{1molH_{2}(g)}{1molZn} = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = <em>190,1L</em>

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL×\frac{101,325}{1atmL} = <em>19,26 kJ</em>

I hope it helps!

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Several different species of birds, including heron, flamingos, and skimmers, can all successfully feed along the coast because
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The correct answer is resource partitioning.  

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When the species differentiate a niche to prevent competition for food resources, it is known as resource partitioning. At certain times, the competition is among the species, known as interspecific competition, and at sometimes it is among the individuals of the similar species, that is, intraspecific competition.  


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