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Lana71 [14]
3 years ago
14

While conducting a lab experiment, Ali calculated that 1.20 E6 Joules of heat were needed to melt 18.5 kilograms of an unknown s

ubstance at its melting point. What is the latent heat of fusion of the substance
Chemistry
1 answer:
nordsb [41]3 years ago
7 0

Answer:

Latent heat of fusion of the substance is 6.49\times 10^{4}J/kg

Explanation:

Latent heat of fusion denotes amount of energy (heat) per unit mass required to melt a solid material at constant temperature and pressure i.e. at it's melting point

Here amount of heat required = 1.20\times 10^{6}J

Mass of unknown substance being melted = 18.5 kg

So, latent heat of fusion of the substance = (required heat energy to melt)/(mass of the unknown substance) = \frac{1.20\times 10^{6}}{18.5}J/kg=6.49\times 10^{4}J/kg

So, latent heat of fusion of the substance is 6.49\times 10^{4}J/kg

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When the water inside a living cell freezes, the ice crystals damage the cell. The wood frog is a unique creature that can survi
professor190 [17]
But i think its B but i need answer too
4 0
3 years ago
what temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution (specific heat capacity = 3
ivolga24 [154]

Answer:

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

Explanation:

Mass of ethylene glycol = m = 100 g

Specific heat capacity of ethylene glycol = c = 3.5 J/g°C

Change in temperature of ethylene glycol = ΔT

Heat loss by the ethylene glycol = Q = 350 J

Q=mc\Delta T

\Delta T=\frac{Q}{mc}=\frac{350 J}{100 g\times 3.5 J/g^oC}

ΔT = 1°C

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

3 0
3 years ago
A given mass of air has a volume of 8.00 L at 60.0°C. At constant pressure, the temperature is increased to 80.0°C. Calculate th
AleksandrR [38]

Answer:

D. 6.00 L

Explanation:

What we have here is an example of Boyle's Law. The equation here is P₁ · V₁ = P₂ · V₂. We know all of the values except for V₂.

60(8) = 80V

<em>Multiply 60 by 8 to get 480.</em>

480 = 80V

<em>Divide both sides by 80.</em>

480/80 = V

6 = V

The final volume for the gas is 6.00 L.

5 0
2 years ago
Liquid hexane
maks197457 [2]

<u>Answer:</u> The mass of H_2O produced is 2.52 g

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

  • <u>For hexane:</u>

Given mass of hexane = 1.72 g

Molar mass of hexane = 86.18 g/mol

Putting values in equation 1, we get:

\text{Moles of hexane}=\frac{1.72g}{86.18g/mol}=0.020mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 8.0 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{8.0g}{32g/mol}=0.25mol

The chemical equation for the combustion of hexane follows:

2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O

By stoichiometry of the reaction:

If 2 moles of hexane reacts with 19 moles of oxygen gas

So, 0.020 moles of hexane will react with = \frac{19}{2}\times 0.020=0.19mol of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hexane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of hexane produces 14 moles of H_2O

So, 0.020 moles of hexane will produce = \frac{14}{2}\times 0.020=0.14mol of H_2O

We know, molar mass of H_2O = 18 g/mol

Putting values in above equation, we get:

\text{Mass of }H_2O=(0.14mol\times 18g/mol)=2.52g

Hence, the mass of H_2O produced is 2.52 g

4 0
3 years ago
A 65.0-gram sample of some unknown metal at 100.0° C is added to 100.8 grams of water at 22.0° C. The temperature of the water r
vampirchik [111]
We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)

The specific heat of the metal is 0.44 J/ (°C × g)
5 0
3 years ago
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