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3 years ago

How many particles are in 2.50 moles of iron (II) chloride?

Chemistry

1 answer:

liq [111]3 years ago

8 0

Answer:

<h2>1.505 × 10²⁴ particles</h2>

Explanation:

The number of particles in iron (II) chloride can be found by using the formula

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 2.5 × 6.02 × 10²³

We have the final answer as

<h3>1.505 × 10²⁴ particles</h3>

Hope this helps you

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Name three sources of atmospheric carbon dioxide.

Marrrta [24]

Decomposition when an organism dies, ocean release and respiration

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3 years ago

When beryllium ion combines with carbonate ion, the numbers of each ion are??

Katyanochek1 [597]

Answer:

One of each

Explanation:

Be is in Group 2, so it loses its two valence electrons in a reaction to form Be²⁺ ions.

Carbonate ion has the formula CO₃²⁻.

We can use the criss-cross method to work out the formula of beryllium carbonate.

The steps are

Write the symbols of the anion and cation.

Criss-cross the numbers of the charges to become the subscripts of the other ion.

Write the formula with the new subscripts.

Divide the subscripts by their highest common factor.

Omit all subscripts that are 1.

When you use this method with Be²⁺ and CO₃²⁻, you might be tempted to write the formula for the beryllium carbonate as Be₂(CO₃)₂

However, you can divide the subscripts by their largest common factor (2).

This gives you the formula Be₁(CO₃)₁.

We omit subscripts that are 1, so the correct formula is

BeCO₃

There is one Be²⁺ ion and one CO₃²⁻ ion in a formula unit of beryllium carbonate.

4 0

4 years ago

How many grams of naoh would react with 507 g fecl2 in the reaction fecl2 + 2naoh fe(oh)2(s) + 2nacl?

rosijanka [135]

<span>Answer: Correct answer is 507g FeCl2 x (1 mol FeCl2 / 126.8 g FeC2) x (1 mol Fe(OH)2 / 1 mol FeCl2) x (89.8 g Fe(OH)2/ 1 mol Fe(OH)2) = 359 g Fe(OH)2.</span>

7 0

3 years ago

Read 2 more answers

When 231. mg of a certain molecular compound X are dissolved in 65. g of benzene (CH), the freezing point of the solution is mea

Elan Coil [88]

Answer:

Molar mass X = 18.2 g/mol

Explanation:

Step 1: Data given

Mass of compound X = 231 mg = 0.231 grams

Mass of benzene = 65.0 grams

The freezing point of the solution is measured to be 4.5 °C.

Freezing point of pure benzene = 5.5 °C

The freezing point constant for benzene is 5.12 °C/m

Step 2: Calculate molality

ΔT = i*Kf*m

⇒ΔT = the freezing point depression = 5.5 °C - 4.5 °C = 1.0 °C

⇒i = the Van't hoff factor = 1

⇒Kf = the freezing point depression constant for benzene = 5.12 °C/m

⇒m = the molality = moles X / mass benzene

m = 1.0 / 5.12 °C/m

m = 0.1953 molal

Step 3: Calculate moles X

Moles X = molality * mass benzene

Moles X = 0.1953 molal * 0.065 kg

Moles X = 0.0127 moles

Step 4: Calculate molar mass X

Molar mass X = mass / moles

Molar mass X = 0.231 grams / 0.0127 moles

Molar mass X = 18.2 g/mol

5 0

3 years ago

Which of the following is a possible Noble gas notation? [Xe] 6s26p6 [Kr] 5s25d85p6 [Xe] 6s25d106p4 [Kr] 5s24d105p6

Mice21 [21]

Answer is: [Kr] 5s24d105p6.

1) [Xe] 6s²6p⁶ is not correct, because element with atomic number 62 (54 + 8) is samarium with electron configuration [Xe] 6s²4f⁶.

Xenon (Xe) is noble gas with atomic number 54, which means it has 54 protons and 54 electrons.

Electron configuration of xenon atom:

₅₄Xe 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶.

2) [Kr] 5s²5d⁸5p⁶ is not correct, because element with atomic number 52 (36 + 16) is tellurium.

Tellurium (Te) is an element with atomic number 52 (52 protons and 52 electrons), it is in group 16 (chalcogens) and period 5 (principal quantum number n = 5, which means it has 5 electrons shell) of Periodic table.

Electron configuration of tellurium atom:

₅₂Te 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁴ or [Kr] 4d¹⁰5s²5p⁶.

Krypton is a chemical element with symbol Kr and atomic number 36, which means it has 36 protons and 36 electrons.

Electron configuration of krypton atom:

₃₆Kr 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶.

3) [Xe] 6s²5d¹⁰6p⁴ is not correct, because element with atomic number 70 (54 + 16) is ytterbium (Yb) with electron configuration [Xe] 6s²4f¹⁴.

4) [Kr] 5s²4d¹⁰5p⁶ is correct, because element with atomic number 54 (36 + 18) is krypton (Kr) with electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ or [Kr] 5s²4d¹⁰5p⁶.

7 0

3 years ago

Read 2 more answers