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2 years ago

Given the following equivalents, make the following conversion: 1.00 knop = ? knips

Chemistry

1 answer:

Makovka662 [10]2 years ago

8 0

Answer:

$6.4knips$

Explanation:

Hello,

In this case, given the stated equivalences, we can use the following proportional factor in order to compute the required knips:

$knips=1.00knop*\frac{6bippy}{1knop} *\frac{8pringle}{10bippy}* \frac{3blip}{18pringle} *\frac{4clips}{5blips} *\frac{10knip}{1clip} \\\\=6.4knips$

Regards.

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A) Write the word equation for the reaction of barium nitride (Ba3N2) with potassium.

artcher [175]

Answer:

Explanation:

In order to balance it, we need to have the same number of atoms of each element on both sides of the equation. There are two atoms of nitrogen on the left, so we need to put 2 in front of K₃N. Now, we have six atoms of potassium on the right, so we need to put 6 in front of K on the left. Finally, there are three atoms of barium on the left, so we put 3 before Ba on the tight. Which means:

Ba₃N₂ + 6K = 2K₃N + 3Ba

Now, we can do the work. First, we determine the molar mass of each reactant ( from the periodic table). Molar mass of the barium is 137, potassium 39 and nitrogen 14.

Ba₃N₂₂ has molar mass of 3Ba and 2N, which means 3 • 137 + 2 • 14 = 439. That means that one mole of Ba₃N₂ weights 439 grams.

We are given grams of reactants, but in order to find the limiting and the excess reactant, we need to transfer it into moles.

We are given 66.5 grams of Ba₃N₂ and we know that 439 grams equals 1 mole. We want to know how many moles there are in 66.5 grams, so the answer is 66.5 / 439 = 0.15 moles.

Let's do the same for potassium. We are given 29 grams of K and we know that 1 mole has 39 grams. We want to know how many moles of K are there in 29 grams, so the answer is 29 / 39 = 0.74 moles.

We now know that 0.15 moles of Ba₃N₂ reacted with 0.74 moles of K. From the balanced equation we see that 1 mole of Ba₃N₂ reacts with 6 moles of K, so the ratio has to be 1:6.

Now let's find limiting and excess reactant. That means that in our reaction, there are more (or less) of one reactant then needed.

We know that we had 0.15 moles of Ba₃N₂ reacting. Let's pretend we don't know the moles of K and let's see with which amount of K should 0.15 moles of barium nitride react, if the ratio is 1:6.

0.15 moles of Ba₃N₂ : x moles of K = 1:6

x = 0.9 moles of K

So, for the completed reaction we need to have 0.9 moles of K, but we previously calculated that we had 0.74. That means that there is less K then needed, so potassium is our limiting reactant, which obviously means that Ba₃N₂ is our excess reactant.

Now, we need to find how many moles of Ba₃N₂ there needed to be for a completed reaction

x moles of Ba₃N₂ : 0.74 moles of K = 1:6

x = 0.124 moles of Ba₃N₂

So we needed to have 0.124 moles, but we had 0.15 of Ba₃N₂, which is 0.15 - 0.124 = 0.026 moles in excess.

If we want to find how many grams that is, we only multiply it with molar mass of Ba₃N₂:

0.026 • 439 = 11.4 grams

That means that only 66.5 - 11.4 = 55.1 grams of Ba₃N₂ reacted.

3 0

2 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

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2 years ago

When dissolved beryllium chloride reacts with dissolved silver nitrate in water, aqueous beryllium nitrate and silver

lisabon 2012 [21]

Answer:

The balanced chemical equation is :

$BeCl_{2} + 2AgNO_{3}\rightarrow Be(NO_{3})_{2} + 2AgCl$

Explanation:

The chemical equation in which number of atoms in reactants is equal to products is called balanced equation.

Formula of , Beryllium Chloride : $[BeCl_{2}$

Beryllium Nitrate : $BeNO_{3}$

Silver Chloride : $AgCl$

Silver Nitrate : $AgNO_{3}$

When beryllium chloride reacts with dissolved silver nitrate in water , following reaction occur :

$BeCl_{2} + 2AgNO_{3}\rightarrow Be(NO_{3})_{2} + 2AgCl$

The number of atoms in reactant as well as in products are balanced :

Be = 1

Ag = 2

N =2

O = 6

Cl = 2

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2 years ago

In two or more complete sentences explain how to balance the chemical equation and classify its reaction type. ___P4 + ___O2 ⟶ _

Serhud [2]

According to the conversation of mass, mass cannot be created or destroyed. This means whatever is done to one side, must be done to the other.

There are 4 Phosphorus atoms on the left, there must be 4 on the right. To do this, you must multiply the P2O3 by 2 to get 4 Phosphorus atoms and 6 Oxygen atoms. Now to balance the Oxygen atoms, you must multiply the oxygen atoms on the left by 3.

1 P4 + 3 O2 —-> 2 P2O3

Lastly, this equation type is synthesis (combination) because two reactants are becoming a single product.

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2 years ago

Read 2 more answers

What quantity, in moles, of oxygen is consumed when 717.4 kJ of energy is evolved from the combustion of a mixture of H2(g) and

mars1129 [50]

Answer: 2.51 moles of oxygen is consumed

Explanation:

Exothermic reaction : It is a type of chemical reaction where the energy is released into the surrounding. In the exothermic reaction, the energy of reactant are more than the energy of product.

In exothermic reaction, the change in enthalpy is, negative

The balanced chemical reaction is:

$2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H=-285.8kJ$

According to stoichiometry:

When 1 mole of oxygen is consumed , energy released = 285.8 kJ

285.8 kJ of energy is evolved when oxygen consumed = 1 mol

Thus whn 717.4 kJ of energy is evolved when oxygen consumed = $\frac{1}{285.8}\times 717.4=2.51$ moles

2.51 moles of oxygen is consumed

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2 years ago