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Aliun [14]
3 years ago
9

How many liters of a 1.5 M solution can you make if you have .50 mol of KCl?

Chemistry
1 answer:
vaieri [72.5K]3 years ago
7 0

Answer:

The answer to your question is 0.33 liters

Explanation:

Data

Volume = ?

Molarity = 1.5 M

number of moles = 0.5

Formula

Molarity = \frac{number of moles }{volume}

Solve for V

Volume = \frac{number of moles}{molarity}

Substitution

Volume = \frac{0.5}{1.5}

Simplification and result

       Volume = 0.33 l

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A graduated cylinder is filled 25.0 mL with water and a piece of granite is placed in the cylinder, displacing the level to 37.2
kvv77 [185]

The volume  of the granite piece  in Cm³  is 12.2 cm³

 <u><em>calculation</em></u>

volume of granite = (volume of cylinder after placing granite - volume  of  cylinder  before  placing  granite

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convert ml  to Cm³

that is 1 ml = 1 cm³

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<em>by cross  multiplication</em>

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3 0
3 years ago
What is the name of the aluminum ion? Al–1 Al+2 Al–3 Al+3
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Most transition metal form more than one cation but aluminum forms the Al3+ cation only.

5 0
4 years ago
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How many atoms are in 13.97 liters of water vapor at STP
Arisa [49]
<span>Let's </span>assume that water vapor has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10</span>⁻³ m³<span>
n = ?
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 0 °C = 273 K (standard temperature)
<span>
By substitution,
</span>101325 Pa x 13.97x 10</span>⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K<span>
                                          n = 0.624 mol
<span>
Hence, the moles of water vapor at STP is 0.624 mol.

According to the </span></span>Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.
<span>
Hence, number of atoms in water vapor = 0.624 mol x </span>6.022 × 10²³ mol⁻¹
<span>                                                                = 3.758 x 10</span>²³<span>

</span>
5 0
3 years ago
The molar mass of I2 is 253.80 g/mol, and the molar mass of NI3 is 394.71 g/mol. How many moles of I2 will form 3.58 g of NI3?
Nat2105 [25]
To answer the problem above first we need to find the difference of molar mass of NI3 from I2, 394.71 g/mol - 253.80 g/mol = 140.91 g/mol. Knowing the molar mass of the difference of NI3 from I2, in equation mass (g) / moles (mol) = molar mass, then we substitute. 3.58g / moles = 140.91 g/mol.
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4 years ago
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ElenaW [278]

Answer:

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3 years ago
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