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sertanlavr [38]
2 years ago
5

50 POINTSSSSS PLZ HELPPPPPP NEED TO SHOW WORK!!!!

Chemistry
1 answer:
VashaNatasha [74]2 years ago
5 0
(4 mol H2O) x (112 kJ / 3 mol H2O) = 149 kJ 

<span>(14.5 g HCl) / (36.4611 g HCl/mol) x (112 kJ / 3 mol HCl) = 14.9 kJ </span>

<span>(475 kJ) / (181 kJ / 2 mol HgO) x (216.5894 g HgO/mol) = 1137 g HgO </span>

<span>(179 kJ) / (181 kJ / 1 mol O2) x (31.99886 g O2/mol) = 31.6 g O2 </span>

<span>(145 kJ) / (112 kJ / 3 mol Cl2) x (70.9064 g Cl2/mol) = 275 g Cl2 </span>

<span>(14.5 g S2Cl2) / (135.0360 g S2Cl2/mol) x (112 kJ / 1 mol S2Cl2) = 12.0 kJ </span>

<span>CaCO3 + 2 NH3 → CaCN2 + 3 H2O; ∆H = –90.0 kJ </span>
<span>(798 kJ) / (90.0 kJ / 2 mol HN3) x (17.03056 g NH3/mol) = 302 g NH3 </span>

<span>(19.7 g H2O) / (18.01532 g H2O/mol) x (90.0 kJ / 3 mol H2O) = 32.8 kJ</span>
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Answer:

The answer would be Uranium-238

Explanation:

It's the only one that lasts longer than 40,000 years

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Why are nonnative species a threat to biodiversity?
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A. They often use up resources that other organisms need.

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Suppose a substance has a heat of fusion equal to 45 calg and a specific heat of 0.75
Varvara68 [4.7K]

Answer:

The substance will be in liquid state at a temperature of 97.3 °C

Note: The question is incomplete. The complete question is given below :

Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)

Explanation:

1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT

Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C

H = 50 × 0.75 × 3 = 112.5 calories

b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g

H = 50 × 45 = 2250 cal

c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.

Amount of energy left = 5000 - 2362.5 = 2637.5 cal

The remaining energy is used to heat the liquid

H = mcΔT

Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change

2637.5 = 50 × 0.75 x ΔT

ΔT = 2637.5 / ( 50*0.75)

ΔT = 70.3 °C

Final temperature of sample = (70.3 + 27) °C = 97.3 °C

The substance will be in liquid state at a temperature of 97.3 °C

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