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zhenek [66]
2 years ago
11

How long would it take for 1.50 mol of water at 100.0 âc to be converted completely into steam if heat were added at a constant

rate of 20.0 j/s ? express your answer to three significant figures and include the appropriate units?
Chemistry
1 answer:
defon2 years ago
6 0
To convert boiling water to steam, that would involve heat of vaporization. The heat of vaporization for water at atmospheric conditions is: ΔHvap = <span>2260 J/g.

Molar mass of water = 18 g/mol

Q = m</span>ΔHvap = (1.50 mol water)(18 g/mol)(<span>2260 J/g) = 61,020 J

Time = Q/Rate = (61,020 J)(1 s/20 J) = 3051 seconds

In order to express the answer in three significant units, let's convert that to minutes.
Time = 3051 s * 1min/30 s = <em>102 min</em></span>
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Tell me the name please<br>​
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Answer:

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What is missing from the solubility graph shown on the right? A graph with temperature in degrees Celsius ranging from 0 to 100
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Answer:

graph title, label for the y axis, and legend explaining the orange and blue colors

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8 0
2 years ago
Read 2 more answers
1. Calculate how many moles of glycine are in a 130.0-g sample of glycine.2. Calculate the percent nitrogen by mass in glycine.
Alexxx [7]

Answer:

n=1.732mol

\% N=18.7\%

Explanation:

Hello!

In this case, since the molecular formula of glycine is C₂H₅NO₂, we realize that the molar mass is 75.07 g/mol; thus, the moles in 130.0 g of glycine are:

n=130.0g*\frac{1mol}{75.07 g}\\\\ n=1.732mol

Furthermore, we can notice 75.07 grams of glycine contains 14.01 grams of nitrogen; thus, the percent nitrogen turns out:

\% N=\frac{14.01}{75.07}*100\% \\\\\% N=18.7\%

Best regards!

4 0
3 years ago
Mass of NaHCO3 sample (g): 1.20 g
sasho [114]

Answer: (22.98977 g Na/mol) + (1.007947 g H/mol) + (12.01078 g C/mol) + ((15.99943 g O/mol) x 3) = 84.0067 g NaHCO3/mol

9.

(1.20 g NaHCO3) / (84.0067 g NaHCO3/mol) = 0.0143 mol NaHCO3

10.

Supposing the question is asking about "how many moles" of CO2. And supposing the reaction to be something like:

NaHCO3 + H{+} = Na{+} + H2O + CO2

(0.0143 mol NaHCO3) x (1 mol CO2 / 1 mol NaHCO3) = 0.0143 mol CO2 in theory

11.

n = PV / RT = (1 atm) x (0.250 L) / ((0.0821 L atm/K mol) x (298 K)) = 0.0102 mol CO2

12.

(0.0143 mol - 0.0102 mol) / (0.0143 mol) = 0.287 = 28.7%

Explanation:

6 0
3 years ago
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