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Crazy boy [7]
3 years ago
13

The number of hydrogen atoms in 0.050 mol of c3h8o3 is

Chemistry
1 answer:
Natalka [10]3 years ago
6 0

Answer:

            2.40 × 10²³ Atoms of Hydrogen

Solution:

The number of hydrogen atoms in C₃H₈O₃ are calculated as;

As, there are eight hydrogen atoms per molecule of C₃H₈O₃, therefore the number of hydrogen atoms per mole of C₃H₈O₃ are calculated as;

            1 mole C₃H₈O₃ contains  =  6.022 × 10²³ Molecules of C₃H₈O₃

So,

       0.05 mole C₃H₈O₃ will contain  =  X Molecules of C₃H₈O₃

Solving for X,

                     X  =  (0.05 × 6.022 × 10²³) ÷ 1

                     X =  3.011 × 10²² Molecules of C₃H₈O₃

As,

            1 Molecule of C₃H₈O₃ contains  =  8 Atoms of Hydrogen

So,

3.011 × 10²² Molecules of C₃H₈O₃ will contain  =  X Atoms of Hydrogen

Solving for X,

                     X =  (3.011 × 10²² × 8) ÷ 1

                     X  =  2.40 × 10²³ Atoms of Hydrogen


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Convert the following with the correct number of significant figures:
larisa [96]

Answer:

1.78 × 10⁹ μg

Explanation:

We have to convert 1.78 kg to μg.

Step 1: Convert 1.78 kilograms to grams

We will use the conversion factor 1 kg = 10³ g.

1.78 kg × 10³ g/1 kg = 1.78 × 10³ g

Step 2: Convert 1.78 × 10³ grams to micrograms

We will use the conversion factor 1 g = 10⁶ μg.

1.78 × 10³ g × 10⁶ μg/1 g = 1.78 × 10⁹ μg

8 0
2 years ago
Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
love history [14]

Answer:

(a) V_B=11.68L

(b) x_{He}=0.533

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol

Thus, since the final pressure is 3.60 bar, we can write:

P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar

The moles of helium could be computed via solver as:

n_{He}=2.358mol

Or algebraically:

3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol

In such a way, the volume of the compartment B is:

V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\  \\V_B=11.68L

Finally, he mole fraction of He is:

x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533

Regards.

8 0
2 years ago
view each ecosystem at your table. Think about the resources that are necessary in order for each to be successful. List all of
TiliK225 [7]

Answer:

Organisms compete for the resources they need to which are survive- air, water, food, and space.

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Answer:

0.56 liters

Explanation:

First we <u>convert 0.80 grams of O₂ into moles</u>, using its molar mass:

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At STP, 1 mol of any given mass occupies 22.4 L. With that information in mind we <u>calculate the volume that 0.025 moles of O₂ gas would occupy</u>:

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Thus the answer is 0.56 liters.

3 0
2 years ago
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Answer: yes true

Explanation: 1. Toward the middle of a river, water tends to flow fastest; toward the margins of the river it tends to flow slowest. 2. In a meandering river, water will tend to flow fastest along the outside bend of a meander, and slowest on the inside bend.

6 0
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