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2 years ago

Determine the number of atoms in 73.0 grams of calcium, Ca. (The

1 answer:

7 0

# of atoms per mol = Avogadro’s # (6.022 x 10^23)

Number of mols = mass of substance / molar mass

73 g / 40.08 g = 1.8 mols of Ca in 73 grams

1.8 mols x avagadro’s # = 1.1 x 10^24 atoms in 73 grams of Ca

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Passing an electric current through a sample of water (H2O) can cause the water to decompose into hydrogen gas (H2) and oxygen g

Andreas93 [3]

2H2O --> 2H2 + O2
The mole H2O:mole O2 ratio is 2:1
Now determine how many moles of O2 are in 50g: 50g × 1mol/32g = 1.56 moles O2
Since 1 mole of O2 was produced for every 2 moles of H2O, we need 2×O2moles = H2O moles
2×1.56 = 3.13 moles H2O
Finally, convert moles to grams for H2O:
3.13moles × 18g/mol = 56.28 g H2O
D) 56.28

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3 years ago

PLEASE HELP ME WITH PHYSICAL SCIENCE!! PRETTY EASY QUESTION

Umnica [9.8K]

Explanation:

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7 0

2 years ago

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what volume of a 0.138 m potassium hydroxide solution is required to neutralize 26.0 ml of a 0.205 m nitric acid solution?

Law Incorporation [45]

A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is ( $M1V1 = M2V2$ ).

Utilize the titration method of $M1V1 = M2V2$ in view that we're given the concentrations of every compound and the quantity of $KOH$ . Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.

M1 = initial mass
V1= initial volume
M2 = final mass
V2= final volume

$(M1V1 = M2V2)$
(0.138)(V1) = (0.205)x(26.0)
V2=(0.205)x(26.0)\ 0.138
V2 = 47.10 M/L
The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L

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4 0

9 months ago

Use this equation for the following problems: 2NaN3 --> 2Na+3N2

olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

4 0

2 years ago

Every atom has atleast two shells ?

VladimirAG [237]

Nope. Hydrogen has one.

4 0

3 years ago

Read 2 more answers