How many atoms of aluminum (AI) are contained in 3.73 moles of aluminum?

Determine the empirical formula of a

k0ka [10]

Answer:

AuCl

Explanation:

Given parameters:

Mass of Gold = 2.6444g

Mass of Chlorine = 0.476g

Unknown:

Empirical formula = ?

Solution:

Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.

Elements Au Cl

Mass 2.6444 0.476

Molar mass 197 35.5

Number of moles 2.6444/197 0.476/35.5

0.013 0.013

Divide by the

smallest 0.013/0.013 0.013/0.013

1 1

The empirical formula of the compound is AuCl

3 0

2 years ago

Urgent! <br> which element do you think has a greater mass? cadmium or zinc?

Alex Ar [27]

Answer:

zinc

Explanation:

5 0

2 years ago

Read 2 more answers

6. In Reaction A, students are instructed to add no more than 0.25 mL of 15 M nitric acid. What volume of 15 M nitric acid is re

enyata [817]

The given question is incomplete, the complete question is:

In Reaction A, students are instructed to add no more than 0.25 mL of 15 M nitric acid. What volume of 15 M nitric acid is required to react with 0.030 g of copper metal? If 1.0 mL of acid contains approximately 20 drops, how many drops of nitric acid are needed?

Answer:

The correct answer is 0.0629 ml and 1.26 drops.

Explanation:

Based on the given question, the equation is:

Cu + 2HNO₃ (aq) ⇒ Cu(NO₃)₂ + H₂

The mass of copper given is 0.030 grams.

The molecular mass of copper is 63.55 gram per mole. The number of moles can be determined by using the formula, n = weight/molecular mass.

Moles of Cu = 0.030 grams/63.55 grams per mole = 0.000472 moles

Based on the reaction, it is clear that 1 mole of Cu reacts with 2 moles of nitric acid, therefore, the number of moles of nitric acid needed will be,

= 0.000472 mol Cu × 2 mol HNO₃ / 1 mole Cu

= 0.000944 mol HNO₃

The concentration of HNO₃ given is 15 M

Now the volume of HNO₃ required to react with 0.030 grams of copper metal will be,

Volume = 0.000944 mol HNO₃ × 1L/15 mol HNO₃ × 1000 ml/ 1L

= 0.0629 ml.

Based on the given information, if 1 ml of nitric acid comprise 20 drops, therefore, 0.0629 ml of the acid will require the drops,

Number of drops of HNO₃ = 0.0629 ml × 20 drops / 1 ml

= 1.26 drops.

7 0

3 years ago

An igneous rock originally has 3 grams of uranium 238 in it. When dated the rock only contains 1.8 grams. What are the parent an

castortr0y [4]

Answer :

The parent and daughter concentrations (in percentages) is, 60 % and 40 % respectively.

The age of rock is $3.32\times 10^9\text{ years}$

Explanation :

First we have to calculate the parent and daughter concentrations (in percentages).

$\text{Parent concentrations}=\frac{1.8g}{3g}\times 100=60\%$

and,

$\text{Daughter concentrations}=\frac{(3-1.8)g}{3g}\times 100=40\%$

As we know that, the half-life of uranium-238 = $4.5\times 10^9$ years

Now we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{4.5\times 10^9\text{ years}}$

$k=1.54\times 10^{-10}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.54\times 10^{-10}\text{ years}^{-1}$

t = time passed by the sample = ?

a = initial amount of the reactant = 3 g

a - x = amount left after decay process = 1.8 g

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.54\times 10^{-10}}\log\frac{3}{1.8}$

$t=3.32\times 10^9\text{ years}$

Therefore, the age of rock is $3.32\times 10^9\text{ years}$

5 0

3 years ago

An unbalanced equation is shown. In this reaction, 200.0 g of FeS2 is burned in 100.0 g of oxygen, and 55.00 g of Fe2O3 is produ

Crazy boy [7]

<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>

Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.

Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3

To find for the theoretical yield, we first determine the limiting reactant.

100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2

Therefore, the limiting reactant is O2.

Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3

Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%

3 0

2 years ago