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earnstyle [38]
2 years ago
7

Your role playing game party of high fantasy warriors arrives at a blacksmith to get new weapons forged. The blacksmith forges a

longsword with 1.74 kg of iron at 1371 °C. which was cooled down in 8.5 L of water initially at 40 °C. What is the final temperature of the iron longsword when it reaches thermodynamic equilibrium? The specific heat capacity of iron is 0.444 J/g °C.
Chemistry
1 answer:
Drupady [299]2 years ago
6 0

Answer:

THE FINAL TEMPREATURE OF IRON SWORD IS

1331°C

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Which of the following CANNOT be determined by looking at the spectra of a star? *
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Answer:

A:temperature

Explanation:

The temperature cannot be determined by looking at the spectra of the star due to lack of the equipment for its measurement. <em>On the other-hand, the remaining statements like the distance from earth, movement towards or away from earth can be determined.</em>

8 0
3 years ago
What is the volume in liters of 321 g of liquid with a density of 0.84 g/mL
Rasek [7]
Density is weight by volume.   

First.  If you divide the weight by density you can find the volume

Second you must convert the ML in to Liters.

\frac{321 \frac{g}{1}}{0.84 \frac{g}{mL}} = \frac{321(g)(mL)}{0.84g}=\frac{321mL}{0.84}=382.14mL

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\frac{1L}{1000mL}

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4 0
3 years ago
Limiting Reactants—————-
denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

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Number of moles of Mg:

\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}.

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}.

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. 2\times 0.270642 = 0.541284\;\text{mol} of MgO will be produced if O₂ is in excess.

How many moles of MgO will be produced?

0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.

What's the mass of 0.541284 moles of MgO?

Formula mass of MgO:

24.301 + 15.999 = 40.300\;\text{g}\cdot\text{mol}^{-1}.

Mass of 0.541284 moles of MgO:

m = n \cdot M = 0.541284\times 40.300 = 21.8\;\text{g}.

7 0
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will this help ?

Explanation:

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