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3 years ago

Calculate how many grams of o2(g) can be produced from heating 87.4 grams of kclo3(s).

1 answer:

4 0

<span>34.2 grams Lookup the atomic weights of the involved elements Atomic weight potassium = 39.0983 Atomic weight Chlorine = 35.453 Atomic weight Oxygen = 15.999 Molar mass KClO3 = 39.0983 + 35.453 + 3 * 15.999 = 122.5483 g/mol Moles KClO3 = 87.4 g / 122.5483 g/mol = 0.713188188 mol The balanced equation for heating KClO3 is 2 KClO3 = 2 KCl + 3 O2 So 2 moles of KClO3 will break down into 3 moles of oxygen molecules. 0.713188188 mol / 2 * 3 = 1.069782282 mols So we're going to get 1.069782282 moles of oxygen molecules. Since each molecule has 2 atoms, the mass will be 1.069782282 * 2 * 15.999 = 34.23089345 grams Rounding the results to 3 significant figures gives 34.2 grams</span>

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Answer:

No it is not a physics because in this there is about classification of element so it is in chemistry that's why it is not physics.

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2 years ago

Carlos is looking at a picture of a convergent boundary on which a volcanic arc and a trench are identified. Which type of conve

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2 years ago

Read 2 more answers

In a titration experiment, 31.4 mL of 1.120 M HCOOH is neutralized by 16.3 mL of Ba(OH)2. What is the concentration of the Ba(OH

11Alexandr11 [23.1K]

Answer : The concentration of the $Ba(OH)_2$ solution is, 2.16 M

Explanation :

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCOOH$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ba(OH)_2$ .

We are given:

$n_1=1\\M_1=1.120M\\V_1=31.4mL\\n_2=2\\M_2=?\\V_2=16.3mL$

Putting values in above equation, we get:

$1\times 1.120M\times 31.4mL=2\times M_2\times 16.3mL\\\\M_2=2.16M$

Thus, the concentration of the $Ba(OH)_2$ solution is, 2.16 M

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3 years ago

6. How much energy is needed to freeze 88.0 g of silver?

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7,040 energy to freeze the silver :)

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2 years ago

You are balancing an ionic compound. Let's call it: AB.

rewona [7]

Answer:- $A_3B_2$

Explanations:- It is given that the charge for A is +2 and the charge for B is -3. The over all compound is neutral means the over all charge is zero. For making the over all charge zero, we need 3 positive ions and 2 negative ions. This makes a +6 charge for A and -6 charge for B and the over all charge is zero.

Also, if we think about the criss cross then charge of A becomes the subscript of B and the charge of B becomes the subscript of A.

So, the formula of the ionic compound is $A_3B_2$ . In this compound the ratio of A to B is 3:2.

The balanced equation could be shown as:

$3A+2B\rightarrow A_3B_2$

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2 years ago