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koban [17]
3 years ago
9

How many liters will 2.76 mol CO2 occupy? 2.76 mol II

Chemistry
1 answer:
sergeinik [125]3 years ago
6 0

This problem is providing information about the moles of carbon dioxide, 2.76 mol, and asks for the volume this amount takes up, turning out to be 61.8 L according to the Avogadro's law:

<h3>Avogadro's law:</h3><h3 />

In chemistry, gas laws are used to relate the behavior of gases by virtue of the their pressure, volume, temperature and moles; thus several gas laws exist for us to do so, however, we here focus on the Avogadro's law which relates the volume and moles when both temperature and pressure are held constant.

In such a way, since no information on the constant variables is given, we assume the mentioned carbon dioxide is at STP, (0 °C and 1 atm), which means that we can use the following equivalence statement derived from the ideal gas law (PV=nRT):

22.4 L = 1 mol

Hence, we calculate the required volume:

2.76molCO_2*\frac{22.4L\ CO_2}{1molCO_2} =61.8L\ CO_2

Learn more about ideal gases: brainly.com/question/11676583

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A saturated solution of barium fluoride, BaF2BaF2, was prepared by dissolving solid BaF2BaF2 in water. The concentration of Ba2+
babunello [35]

Answer: 1.70\times 10^{-6}

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as

The equation for the ionization of the BaF_2 is given as:

BaF_2\rightarrow Ba^{2+}+2F^-

When the solubility of BaF_2 is S moles/liter, then the solubility of Ba^{2+}  will be S moles/liter and solubility of F^- will be 2S moles/liter.

By stoichiometry of the reaction:

1 mole of BaF_2 gives 1 mole of Ba^{2+ and 2 moles of F^-

K_{sp}=[Ba^{2+}][F^{-}]^2

K_{sp}=s\times (2s)^2

K_{sp}=4\times (7.52\times 10^{-3})^3

K_{sp}=1.70\times 10^{-6}

Thus K_{sp} for BaF_2 is 1.70\times 10^{-6}

7 0
3 years ago
An aqueous solution is 0.467 m in hcl. what is the molality of the solution if the density is 1.23 g/ml?
Reika [66]
Answer:
molarity = 0.385 moles/kg

Explanation:
Assume that the volume of the aqueous solution given is 1 liter = 1000 ml
Now, density can be calculated using the following rule:
density = mass / volume
Therefore:
mass = density * volume = 1.23 * 1000 = 1230 grams
Now, 0.467 m/L * 1L = 0.467 moles of HCl
We will get the mass of the 0.467 moles of HCl as follows:
mass = molar mass * number of moles = (1+35.5)*0.467 = 17.0455 grams
Now, we have the mass of the solution (water + HCl) calculated as 1230 grams and the mass of the HCl calculated as 17.0455 grams. We can use this information to get the mass of water as follows:
mass of water = 1230 - 17.0455 = 1212.9545 grams
Finally, we will get the molarity as follows:
molarity = number of moles of solute / kg of solution
molarity = (0.467) / (1212.9594*10^-3)
molarity = 0.385 mole/kg 

Hope this helps :)

5 0
3 years ago
Read 2 more answers
What is the second most abundant element in stars? A. helium B. iron C. hydrogen D. oxygen
devlian [24]
<span>The </span>abundance of a chemical element<span> is a measure of the </span>occurrence<span> of the </span>element<span> relative to all other elements in a given environment. Abundance is measured in one of three ways: by the </span>mass-fraction<span> (the same as weight fraction); by the </span>mole-fraction<span> (fraction of atoms by numerical count, or sometimes fraction of molecules in gases); or by the </span>volume-fraction<span>. Volume-fraction is a common abundance measure in mixed gases such as planetary atmospheres, and is similar in value to molecular mole-fraction for gas mixtures at relatively low densities and pressures, and </span>ideal gas<span> mixtures. Most abundance values in this article are given as mass-fractions.

</span>
8 0
3 years ago
Matter that has a uniform and definite composition is called?
enot [183]
Matter that has a uniform and definite composition is called a vapor
8 0
3 years ago
Source from which organisms benefit
vampirchik [111]

Answer:

a ecosystem

Explanation:

7 0
4 years ago
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