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3 years ago

Stoichiometry: Calculate the moles of H2 produced by .042 g Mg in the equation Mg + 2HCl > MgCl2 + H2

Chemistry

1 answer:

Evgesh-ka [11]3 years ago

6 0

Answer:

Mg (s) + 2HCI2 (aq) → MgCI2 (aq) + H2 (g)

I think this is the correct answer I not a 100% sure if it is correct.

Explanation:

Guessing

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The Germans were the first to send up a satellite into space.

vladimir2022 [97]

Answer: Thats false the Russians were the first

Explanation:

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3 years ago

Read 2 more answers

Stannum has a body centered tetragonal with lattice constant, a = b = 5.83A and c = 3.18A. If the atomic radius is 0.145 nm, det

Fed [463]

Answer:

the atomic packing factor of Sn is 0.24

Explanation:

a = b = 5.83A and c = 3.18A.

Volume of unit cell = a²c

= (5.83)² * 3.18 * 10⁻²⁴ cm³

= 1.08 * 10⁻²²cm³

Volume of atoms =

$2 \times \frac{4}{3} \pi r^3$

(∴ BCC, effective number of atom is 2)

Volume of atoms =

$2 * \frac{4}{3} *3.14*(0.145*10^-^7cm)^3$

= 2.55*10⁻²³cm³

$\text {Atomic packing factor}=\frac{\text {volume occupied by atom}}{\text {volume of unit cell }}$

$=\frac{2.55*10^-^2^3}{1.08*10^-^2^2} \\\\=0.24$

<h3>therefore, the atomic packing factor of Sn is 0.24</h3>

4 0

4 years ago

How to do dots and cross diagrams

Likurg_2 [28]

Answer:

Draw circles to represent the electron shell of each atom overlapping the circles where the atoms are bonded. Add dots to represent the outer electrons of one type of atom (H). Add crosses to represent the outer electrons of the other type of atom (Cl). Make sure the electrons are always in pairs.

6 0

3 years ago

scandium47 has a half-life of 35s. suppose you have a 45g sample of scadium 47 how much of the sample remains unchanged after 14

Mars2501 [29]

The much of the sample that would remain unchanged after 140 seconds is 2.813 g

Explanation

Half life is time taken for the quantity to reduce to half its original value.

if the half life for Scandium is 35 sec, then the number of half life in 140 seconds

=140 sec/ 35 s = 4 half life

Therefore 45 g after first half life = 45 x1/2 =22.5 g

22.5 g after second half life = 22.5 x 1/2 =11.25 g

11.25 g after third half life = 11.25 x 1/2 = 5.625 g

5.625 after fourth half life = 5.625 x 1/2 = 2.813

therefore 2.813 g of Scandium 47 remains unchanged.

4 0

3 years ago

One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO

zhenek [66]

Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)

The stoichiometric ratio between Na₃(CO₃)(HCO₃)·2H₂O(s) and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
= 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
= 1000 x 10³ g / 226 g mol⁻¹
= 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x $\frac{3}{2}$
= 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
= 703540.02 g
= 703.540 kg

Hence, the theoretical yield of Na₂CO₃ = 703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
= 92.34%

7 0

3 years ago