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3 years ago

How many moles are equal to 1.3 x 1024 atoms of aluminum?

Chemistry

1 answer:

marshall27 [118]3 years ago

8 0

Answer:

<h3>The answer is 2.16 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

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

$n = \frac{1.3 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 2.159468...$

We have the final answer as

<h3>2.16 moles</h3>

Hope this helps you

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8KCIO3 +1C12H22011

docker41 [41]

3.2 g KClO3

Explanation:

1.1 g C12H22O11 × (1 mol C12H22O11/342.3 g C12H22O11)

= 0.0032 mol C12H22O11

0.0032 mol C12H22O11 × (8 mol KClO3/1 mol C12H22O11)

= 0.026 mol KClO3

Therefore, the minimum amount of KClO3 needed is

0.026 mol KClO3 × (122.55 g KClO3/1 mol KClO3)

= 3.2 g KClO3

5 0

3 years ago

A solution contains 1 LaTeX: \times\:×10−4 M OH– ions. Calculate the solution pH value, and determine if the solution is acidic,

docker41 [41]

Answer:

pH = 10

The solution is basic.

Explanation:

A solution contains 1 × 10⁻⁴ M OH⁻ ions. First, we will calculate the pOH.

pOH = -log [OH⁻]

pOH = -log 1 × 10⁻⁴

pOH = 4

We can find the pH of the solution using the following expression.

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 4 = 10

Since the pH > 7, the solution is basic.

5 0

4 years ago

The reaction 3 BrO- (aq) --> BrO3- (aq) + 2 Br - (aq) in basic solution is second order in BrO-(aq) with a rate cons

Phoenix [80]

Answer:

0.124 M

Explanation:

The reaction obeys second-order kinetics:

$r = k[BrO^-]^2$

According to the integrated second-order rate law, we may rewrite the rate law in terms of:

$\dfrac{1}{[BrO^-]_t} = kt + \dfrac{1}{[BrO^-]_o}$

Here:

$k$ is a rate constant,

$[BrO^-]_t$ is the molarity of the reactant at time t,

$[BrO^-]_o$ is the initial molarity of the reactant.

Converting the time into seconds (since the rate constant has seconds in its units), we obtain:

$t = 1.00 min = 60.0 s$

Rearranging the integrated equation for the amount at time t:

$[BrO^-]_t = \dfrac{1}{kt + \dfrac{1}{[BrO^-]_o}}$

We may now substitute the data:

$[BrO^-]_t = \dfrac{1}{0.056 M^{-1}s^{-1}\cdot 60.0 s + \dfrac{1}{0.212 M}} = 0.124 M$

7 0

3 years ago

Using the following equation 5KNO2+2KMnO4+3H2SO4=5KNO3+2MnSO4+K2SO4+3H20. Starting with 2.47 grams of KNO2 and excess KMnO4 how

Aleonysh [2.5K]

Given equation : $5KNO_{2} + 2KMnO_{4} + 3H_{2}SO_{4}\rightarrow 5KNO_{3} + 2MnSO_{4} + K_{2}SO_{4} + 3H_{2}O$

Given information = 2.47 grams KNO2 and excess KMnO4 and we need to find grams of water (H2O).

Since KMnO4 is in excess, so grams of water(H2O) can be calculated using grams of KNO2 with the help of stoichiometry.

To find grams of water(H2O) from grams of KNO2 , we need to follow three steps.

Step 1. Convert 2.47 grams of KNO2 to moles of KNO2.

$Moles = \frac{grams}{molar mass}$

Molar mass of KNO2 = 85.10 g/mol

$Moles = 2.47 gram KNO2\times \frac{1 mol KNO2}{85.10 gram KNO2}$

Moles = 0.0290 mol KNO2

Step 2. Convert moles of KNO2 to moles of H2O using mole ratio.

Mole ratio are the coefficient present in front of the compound in the balanced equation.

Mole ratio of KNO2 : H2O is 5 : 3 (5 coefficient of KNO2 and 3 coefficient of H2O)

$0.0290 mol KNO2\times \frac{3 mol H2O}{5 mol KNO2}$

Mole = 0.0174 mol H2O

Step 3. Convert mole of H2O to grams of H2O

Grams = Moles X molar mass

Molar mass of H2O = 18.00 g/mol

$Grams = 0.0174 mol H2O\times \frac{18 g H2O}{1 mol H2O}$

Grams of water = 0.313 grams H2O

Summary : The above three steps can also be done in a singe setup as shown below.

$2.47 gram KNO2\times \frac{(1 mol KNO2)}{(85.10 gram KNO2)}\times \frac{(3 mol H2O)}{(5 mol KNO2)}\times \frac{(18.00 gram H2O)}{(1mol H2O)}$

In the above setup similar units get cancelled out and we will get grams of H2O as 0.313 grams water (H2O)

8 0

4 years ago

Which of the following is best revealed by the Bohr Model? .

valentina_108 [34]

The best answer among the following choices would be the fourth option D).

3 0

3 years ago