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sweet [91]
4 years ago
10

How many moles are contained in 3.131 × 1024 particles?

Chemistry
2 answers:
KiRa [710]4 years ago
4 0
All are wrong, it would be 5.324e-21 moles
AleksAgata [21]4 years ago
3 0

<u>Given:</u>

The number of particles = 3.131 * 10²⁴

<u>To determine:</u>

The number of moles corresponding to the given particles

<u>Explanation:</u>

1 mole corresponds to Avogadro's number of particles = 6.023 * 10²³ particles

Therefore, the given particles would correspond to:

3.131* 10²⁴ particles * 1 moles/6.023 * 10²³ particles = 5.199 moles

Ans: A) 5.199 moles are contained in 3.131* 10²⁴ particles



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A Starting substance in a chemical reaction is called a _______________.
djverab [1.8K]

Answer:

1.)REACTANT

  • These starting substances of a chemical reaction are called the <em>reactants</em>, and the new substances that result are called the products.
5 0
3 years ago
A higher amplitude means...?
vekshin1

Answer:

High amplitude is equivalent to loud sounds.

Explanation:

Larger the amplitude, the higher the energy. In sound, amplitude refers to the magnitude of compression and expansion experienced by the medium the sound wave is travelling through. This amplitude is perceived by our ears as loudness.

4 0
3 years ago
Calculate the mass, in grams, of cucl2 (mw = 134.452 g/mol) required to prepare 250.0 ml of a 6.11 % w/v cu2 (mw = 63.546 g/mol)
Sever21 [200]

6.11% w/v of Cu2+ implies that 6.11 g of Cu2+ is present in 100 ml of the solution

therefore,  250 ml of the solution would have: 250 ml * 6.11 g/100 ml = 15.275 g

# moles of Cu2+ = 15.275 g/63.546 g mole-1 = 0.2404 moles

1 mole of CuCl2 contain 1 mole of Cu2+ ion

Hence, 0.2404 moles of Cu2+ would correspond to 0.2404 moles of CuCl2

Molar mass of CuCl2 = 134.452 g/mole

The mass of CuCl2 required = 0.2404 moles * 134.452 g/mole = 32.32 grams

6 0
3 years ago
What is the specific heat of a substance if 1450 calories are required to raise the temperature of a 240g sample by 20℃?
bazaltina [42]

Answer:

6960 J/kg°C

Explanation:

specific heat= mass×specific heat capacity×increase in temperature

specific heat= 0.240×1450×20= 6960 J/kg°C

hope it helps!

5 0
3 years ago
Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by pheno
melamori03 [73]

1.062 mol/kg.

<em>Step 1</em>. Write the balanced equation for the neutralization.

MM = 204.22 40.00

KHC8H4O4 + NaOH → KNaC8H4O4 + H2O

<em>Step 2</em>. Calculate the moles of potassium hydrogen phthalate (KHP)

Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)

= 4.035 mmol KHP

<em>Step 3</em>. Calculate the moles of NaOH

Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)

= 4.035 mmol NaOH

<em>Step 4</em>. Calculate the mass of the NaOH

Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)

= 161 mg NaOH

<em>Step 5</em>. Calculate the mass of the water

Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g

= 37.973 g

<em>Step 6</em>. Calculate the molal concentration of the NaOH

<em>b</em> = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg

3 0
4 years ago
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