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3 years ago

How many atoms are in 0.0246 mol K?

Chemistry

1 answer:

Alexxx [7]3 years ago

3 0

Answer:

There are 6.022 × 1023 atoms of potassium in every mole of potassium. Since one mole of KOH contains one mole of K, the answer is 6.022×1023 atoms of K.

Explanation:

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The formula for bismuth(III) arsenide is..?

vlabodo [156]

Answer:

BiAs

Explanation:

6 0

3 years ago

Be sure to answer all parts.

tangare [24]

Answer:

(a) cesium bromide (CsBr): 9.15 grams

(b) calcium sulfate (CaSO4): 5.85 grams

(d) lithium dichromate (Li2Cr2O7): 9.88 grams

(e) potassium oxalate (K2C2O4): 7.15 grams

Explanation:

(a) cesium bromide (CsBr):

Molar mass of CsBR = 212.81 g/mol

Number of moles = molarity * volume

Number of moles = 0.100 M *0.43 L

Number of moles = 0.043 moles

Mass of CsBr required = moles * Molar mass

Mass of CsBr required = 0.043 moles * 212.81 g/mol

Mass of CsBr required = 9.15 grams

(b) calcium sulfate (CaSO4):

Molar mass of CaSO4 = 136.14 g/mol

Mass of CaSO4 required = moles * Molar mass

Mass of CaSO4 required = 0.043 moles * 136.14 g/mol

Mass of CaSO4 required = 5.85 grams

(c) sodium phosphate (Na3PO4):

Molar mass of Na3PO4 = 163.94 g/mol

Mass of Na3PO4 required = moles * Molar mass

Mass of Na3PO4 required = 0.043 moles * 163.94 g/mol

Mass of Na3PO4 required = 7.05 grams

(d) lithium dichromate (Li2Cr2O7):

Molar mass of Li2Cr2O7 = 229.87 g/mol

Mass of Li2Cr2O7 required = moles * Molar mass

Mass of Li2Cr2O7 required = 0.043 moles * 229.87 g/mol

Mass of Li2Cr2O7 required = 9.88 grams

(e) potassium oxalate (K2C2O4):

Molar mass of K2C2O4 = 166.22 g/mol

Mass of K2C2O4 required = moles * Molar mass

Mass of K2C2O4 required = 0.043 moles * 166.22 g/mol

Mass of K2C2O4 required = 7.15 grams

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3 years ago

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kotykmax [81]

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3 years ago

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What is the temperature of 0.750 mol of a gas stored in a 6,050 mL cylinder al 221 atm?

Rasek [7]

Answer:

T = 246 K

Explanation:

Given that,

Number of moles, n = 0.750 mol

The volume of the cylinder, V = 6850 mL = 6.85 L

Pressure of the gas, P = 2.21 atm

We need to find the temperature of the gas stored in the cylinder. We know that,

PV= nRT

Where

R is gas constant

T is temperature

So,

$T=\dfrac{PV}{nR}\\\\T=\dfrac{2.21\times 6.85}{0.75\times 0.0821}\\T=245.85\ K$

T = 246 K

So, the temperature of the gas is equal to 246 K.

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2 years ago

Yusef adds all of the values in his data set and then divides by the number of values in the set. What is Yusef most likely find

Zanzabum

He is most likely finding the mean of the data.

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3 years ago

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