Please help me. Quickly!!!

tia_tia [17]

Vocabulary. Balanced chemical equation: A chemical equation in which the number of each type of atom is equal on the two sides of the equation.

Hope I helped! (´▽`)

__________________________________________________________

単語。平衡化学反応式：各タイプの原子の数が方程式の両側で等しい化学反応式。

私が助けてくれたらいいのに！(´▽`)

5 0

3 years ago

Read 2 more answers

Directions are in the picture

Mrrafil [7]

Answer:

Question Clear

Explanation:

Make sure your question is clear.

6 0

2 years ago

If 125g of KClO3 is heated, what is the total mass of the products?

Andrej [43]

Given parameters:

Mass of KClO₃ = 125g

Unknown:

Total mass of the products = ?

When KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

2KClO₃ → 2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

Find the number of moles of the given reactant:

Number of moles = $\frac{mass}{molar mass}$

molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5g/mol

So number of moles of KClO₃ = $\frac{125}{122.5}$ = 1.02moles

2. Now, using this number of moles, find the number of moles of the products using this value;

2 moles of KClO₃ produced 2 moles of KCl

1.02 moles of KClO₃ will also produce 1.02moles of KCl

2 moles of KClO₃ produced 3 moles of O₂

1.02 moles of KClO₃ will produce $\frac{1.02 x 3} {2}$ mole = 1.53 moles of O₂

3. Now find the masses of each product;

Mass = number of moles x molar mass

molar mass of KCl = 39 + 35.5 = 74.5g/mol

molar mass of O₂ = 16 x 2 = 32g/mol

Mass of KCl = 74.5 x 1.02 = 75.99g

Mass of O₂ = 32 x 1.53 = 48.96g

Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

= 124.95g

This value is approximately the same as that of mass of KClO₃

7 0

3 years ago

Can anyone help me on this please?

Bad White [126]

This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them. This is not an easy question.

The idea is that the "half-life" is a certain amount of time. It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.

Look over the table. There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds. Gosh ... 2,250 is exactly half of 4,500 ! So the length of time from 12,000 seconds until ' y ' is the half life of this substance ! But how can we find the length of the half-life ? ? ?

Maybe we can figure it out from other information in the table !

Here's what I found:

Do you see the time when there were 3,600 atoms of it ?
That's 20,000 seconds.

... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left.

==> 450 is in the table ! That's at 95,000 seconds.

So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.

The length of time is (95,000 - 20,000) = 75,000 sec

                               3 half lifes = 75,000 sec

Divide each side by 3 :   1 half life = 25,000 seconds

There it is ! THAT's the number we need. We can answer the question now.

==> 2,250 atoms is half of 4,500 atoms.

==> ' y ' is one half-life later than 12,000 seconds

==> ' y ' = 12,000 + 25,000

         y   = 37,000 seconds .

Check:
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.

As I said earlier, this is not the simplest half-life problem I've seen.
You really have to know what you're doing on this one. You can't
bluff through it.

7 0

3 years ago

Question 2 of 50

wolverine [178]

The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

Write the balanced chemical equation for the given reaction.
Convert the mass of calcium carbonate into moles.
Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

$CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}$

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

$mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}$

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

$mol \ CaO \ = 0.2498 \ mol$

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

$mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g$

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

$\boxed {mass \ CaO \ = 14 \ g}$

Learn More

Learn more about stoichiometry brainly.com/question/12979299
Learn more about mole conversion brainly.com/question/12972204
Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

5 0

3 years ago