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2 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Chemistry

1 answer:

jekas [21]2 years ago

4 0

Answer:

Abiotic is non living thing, while biotic is a living thing

hope this info helps

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How many atoms of potassium and how

Varvara68 [4.7K]

KOH? 6.022 x 10^23 atoms of potassium(K) are in one mole of KOH

8 0

3 years ago

Part 1. Determine the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm. Show your work.

zaharov [31]

If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.

An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given :

V = 2.4 L = 0.0024

P = 86126.25 Pa

T = 287 K

m = 0.622

R = 8.314

The ideal gas equation is given below.

n = PV/RT

n = 86126.25 x 0.0024 / 8.314 x 287

n = 0.622 / molar mass (n = Avogardos number)

Molar mass = 7.18 g

Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g

More about the ideal gas equation link is given below.

brainly.com/question/4147359

#SPJ1

4 0

1 year ago

The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

$\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

$Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)$

3s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=(3s)^3\times s$

We are given:

$s=2.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}$

Hence, the solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

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3 years ago

Which of the three components of the atom has the smallest mass?

Lesechka [4]

Electron has the smallest mass

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3 years ago

Select the correct answer.

vaieri [72.5K]

Answer:

Explanation:

compounds ..........

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3 years ago