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2 years ago

0.159 mol 2.25 M = A L of HCI

Chemistry

2 answers:

mylen [45]2 years ago

4 0

Answer will be
0.159 mol
2.25 M
= A L of HCI
.0707

Hope this helps.

Verizon [17]2 years ago

3 0

Answer:

hmmm?what?

Explanation:

ayusin mo flece:>

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What is the molecular formula for water?

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2 years ago

Read 2 more answers

One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are remo

frozen [14]

Answer:

The total work is 4957.45J

Explanation:

For an ideal gas, at constant temperature the definition of work (W) is

$W = - P.dV = - n.R.T \int\limits^i_f {\frac{dV}{V} \\W = - n.R.T. Ln (\frac{V_{f}}{V_{I}})\\W = - n.R.T. Ln (\frac{P_{i}}{P_{f}})$

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

$W_{1} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{5.50atm}{2.43atm}) = 23.44atm.Lx101.325\frac{J}{atm.L} =2375.44J$

Stape 2: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

$W_{2} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{2.43atm}{1.00atm}) = 25.48atm.Lx101.325\frac{J}{atm.L} =2582.01J$

The total work is the sum of the two steps

$W = W_{1} + W_{2} = 2375.44J + 2582.01J = 4957.45J$

4 0

3 years ago

Determine the quantity of bromine atoms in 18.0g of HgBr2 (using factor label method)

Gnoma [55]

Considering the definition of molar mass and Avogadro's Number, the quantity of bromine atoms in 18.0g of HgBr₂ is 6.023×10²² atoms.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

The molar mass of a compound (also called Mass or Molecular Weight) is the sum of the molar mass of the elements that form it (whose value is found in the periodic table) multiplied by the number of times they appear in the compound.

<h3>Molar mass of HgBr₂</h3>

In this case, you know the molar mass of the elements is:

Hg= 200 g/mole
Br= 79.9 g/mole

So, the molar mass of the compound HgBr₂ is calculated as:

HgBr₂= 200 g/mole + 2×79.9 g/mole

Solving:

HgBr₂= 359.98 g/mole

Definition of Avogadro's Number

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>Quantity of bromine atoms in 18.0g of HgBr₂</h3>

Factor-label describes a technique to convert one quantity to another quantity. This method uses the fact that any number or expression can be multiplied by one without changing its value; this is, consists in multiply the value you are given, unit of measurement included, by the conversion factor in fraction form.

Finally, knowing that 2 moles of Br are present in 1 mole of HgBr₂, and considering the molar mass of HgBr₂ and the definition on Avogadro's number, the quantity of bromine atoms in 18.0g of HgBr₂ is calculated as:

$18 gr HgBr_{2}\frac{1 mole HgBr_{2}}{359.98 gHgBr_{2}} \frac{2 mole Br}{1 mole HgBr_{2} } \frac{6.023x10^{23}atoms Br }{1 mole Br} =6.023x10^{22}atoms$

In summary, the quantity of bromine atoms in 18.0g of HgBr₂ is 6.023×10²² atoms.

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molar mass:

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Avogadro's Number:

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<u>brainly.com/question/1528951</u>

8 0

2 years ago