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leonid [27]
4 years ago
6

NaCl has a mc020-1.jpgHfus = 30.2 kJ/mol. What is the mass of a sample of NaCl that needs 732.6 kJ of heat to melt completely?

Chemistry
2 answers:
Sidana [21]4 years ago
8 0

Answer:

d.) 1417.7   on edge 2020

Explanation:

hope this helps!

lilavasa [31]4 years ago
5 0

Answer: The mass of the sample will be 1417.7 grams.

Explanation:

We are given:

\Delta H_{fusion}=30.2kJ/mol

This means that 1 mole of NaCl has an enthalpy of fusion of 30.2 kJ

1 mole of NaCl has a mass of 58.44 grams.

So, 30.2 kJ of heat is require for a mass 58.44 grams of NaCl

So, 732.6 kJ of heat will be required for = \frac{58.44g}{30.2kJ}\times 732.6 kJ = 1417.65 grams of NaCl.

Hence, the mass of NaCl sample will be 1417.7 grams.

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A 32.8 g iron rod, initially at 22.4 C, is submerged into an unknown mass of water at 63.1 C, in an insulated container. The fin
laila [671]

Answer:

mass water = 32.4 g

Explanation:

specific heat iron = 0.450 J/g°C

specific heat water = 4.18 J/g°C

32.8 x 0.450 ( 59.1 - 22.4) + mass water x 4.18 ( 59.1- 63.1)=0

541.7 - mass water x 16.7 = 0

mass water = 32.4 g

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What is true regarding the calculation in redox titration? group of answer choices moles of the oxidant equal mole of the reduct
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Moles of titrant are lost in the calculation in redox titration.

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4 0
2 years ago
How does a refracting telescope work?
Yakvenalex [24]

Answer:

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Explanation:

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7 0
3 years ago
A certain protein was found to contain 0.292% manganese by mass. Determine the minimum molecular mass of the protein.
Lisa [10]
55u \ \ \ \ \ \ \Rightarrow \ \ \ \ \ 0,292\%\\
x \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ 100\%\\\\
x = \frac{55u*100\%}{0,292\%}\approx18835,62u
4 0
3 years ago
If 2.0×10−4 moles of S2O2−8 in 170 mL of solution is consumed in 170 seconds , what is the rate of consumption of S2O2−8?
Ostrovityanka [42]

Answer : The rate of consumption of S_2O_2^{-8} is, 7.0\times 10^{-6}M/s

Explanation : Given,

Moles of S_2O_2^{-8} = 2.0\times 10^{-4}mol

Volume of solution = 170 mL = 0.170 L     (1 L = 1000 mL)

Time = 170 s

First we have to calculate the concentration.

\text{Concentration}=\frac{\text{Moles}}{\text{Volume of solution}}

\text{Concentration}=\frac{2.0\times 10^{-4}mol}{0.170L}

\text{Concentration}=1.2\times 10^{-3}M

Now we have to calculate the rate of consumption.

\text{Rate of consumption}=\frac{\text{Concentration}}{\text{Time}}

\text{Rate of consumption}=\frac{1.2\times 10^{-3}M}{170s}

\text{Rate of consumption}=7.0\times 10^{-6}M/s

Thus, the rate of consumption of S_2O_2^{-8} is, 7.0\times 10^{-6}M/s

8 0
3 years ago
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