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Schach [20]
2 years ago
13

For the reaction, calculate how many moles of the product form when 0.046 mol of O2 completely reacts.

Chemistry
1 answer:
aleksandrvk [35]2 years ago
3 0

Answer: 0.023 moles of CaO.

Explanation:

The balanced equation tells us that we'll get 2 moles of CaO for every one mole of O2.  Since we have 0.046 moles of O2, we'll get 0.023 moles of CaO.  (2 sig figs)

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An inflatable toy starts with 1.05 moles of air and a volume of 5.17 liters. When fully inflated, the volume is 8.00 liters. If
padilas [110]

Answer:-  1.62 moles

Solution:- At constant temperature and pressure, volume is directly proportional to the moles of the gas.

\frac{V_1}{V_2}=\frac{n_1}{n_2}

from given data, V_1 = 5.17 L, n_1 = 1.05 moles

V_2 = 8.00 L, n_2 = ?

Let's plug in the values in the formula:

\frac{5.17L}{8.00L}=\frac{1.05moles}{n_2}

On cross multiply:

n_2=\frac{8.00L(1.05moles)}{5.17L}

n_2 = 1.62 moles

So, now the toy contains 1.62 moles of the air.


5 0
3 years ago
Which statement accurately compares the trends in atomic number and atomic mass in the periodic table
Tresset [83]

Answer: is C

Explanation:

Both the atomic mass and the atomic number increase from left to right

6 0
3 years ago
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What three factors affect the pressure of a gas in a closed container?
maks197457 [2]
Number of moles of the gas, Temperature and the volume of the gas.
4 0
2 years ago
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What is true about all chemical equations?<br> Links are reported
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Answer:Your answer is A

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6 0
3 years ago
What are the molarity and osmolarity of a 1-liter solution that contains half a mole of calcium chloride? How many molecules of
givi [52]

Answer:

Molarity = 0.5 M

Osmolarity = 0.5 x 2 = 1 Osmpl.

Molecules of Cl2 = 6.02 x 10^{23} / 4= 1.505 x 10^{23} no. of molecules

Explanation:

If we add half mole in 1L volume than molarity will obviously be 0.5 M.

The osmolarity is molarity multiplies by number of dissociates of solute that for CaCl2 are 2. So, 2 x 0.5 = 1

Half will be molecules of Ca and half will be of Cl2 for 0.5M.

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3 years ago
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