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shutvik [7]
2 years ago
8

(PLEASE HELP DUE TODAY!) A gas has a volume of 1000.0 mL at a temperature of 20.OK and a pressure of 1.0 atm. What will be the n

ew volume when the temperature is changed to 40.0K and the pressure is changed to
0.5 atm?
Chemistry
1 answer:
Nataliya [291]2 years ago
7 0

Answer:

4 L or 4000 mL

Explanation:

This question asks for the use of combined gas law:

(P1)(V1) / (T1) = (P2)(V2) / (T2)

...where pressure (P) is in atm, volume (V) is in liters, and temperature (T) is in kelvin.

1000 ml is 1 L, and everything else is already in the right units.

(1 atm)(1 L) / (20 K) = (0.5 atm)(V2) / (40 K)

Just solve the equation for V2, and you get a new volume of 4 liters

(1)(1) / (20) = (.5x) / (40)

1 / 20 = .5x / 40

2 = .5x

x = 4

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How many grams of oxygen gas must react to give 2.10g of ZnO?
Advocard [28]
The chemical reaction is written as:

2Zn + O2 = 2ZnO

We are given the amount of the product to be produced from the reaction. We use this value and the relation of the substances in the reaction to calculate what is asked. We do as follows:

2.10 g ZnO ( 1 mol / 81.408 g ) ( 1  mol O2 / 2 mol ZnO ) ( 32 g / 1 mol ) = 0.414 g O2 is needed
8 0
3 years ago
What is the main difference between a generator and an electric motor? _______
garri49 [273]

Answer:

A generator produces electricity and an electric motor consumes electricity

4 0
3 years ago
Read 2 more answers
The combustion of butane produces heat according to the equation 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l), ΔH°rxn= –5,314 kJ/mol
Dafna11 [192]

Answer:

665 g

Explanation:

Let's consider the following thermochemical equation.

2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol

According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:

-1.00 × 10⁴ kJ × (8 mol CO₂/-5,314 kJ) = 15.1 mol CO₂

The molar mass of CO₂ is 44.01 g/mol. The mass corresponding to 15.1 moles is:

15.1 mol × 44.01 g/mol = 665 g

4 0
3 years ago
3 points
VLD [36.1K]

Answer:

<h3>The answer is 8.29 %</h3>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

actual density = 19.30g/L

error = 20.9 - 19.3 = 1.6

We have

p(\%) =  \frac{1.6}{19.3}  \times 100 \\  = 8.290155440...

We have the final answer as

<h3>8.29 %</h3>

Hope this helps you

5 0
3 years ago
Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou
Ulleksa [173]

Answer:

4.36~g~XY

Explanation:

In this case, we can start with the reaction:

2X + Y_2~->~2XY

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:

3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X

4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2

Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):

\frac{0.04~mol~X}{1}=~0.04

\frac{0.0875~mol~Y_2}{2}=0.04375

The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol Y_2 = 48 g Y_2 (therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY

I hope it helps!

6 0
3 years ago
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