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2 years ago

If 45.8 grams of potassium chlorate decomposes, how many grams of oxygen gas can be produced? 2KClO3 → 2KCl + 3O2

1 answer:

3 0

To do this problem, we must first look at the balanced chemical equation for the decomposition of potassium chlorate:

2KClO3 --> 2KCl + 3O2 

We can take the given amount of grams, and use the molar mass of KClO3 to convert to moles. Then, we can use the stoichiometric ratios to relate moles of KClO3 to moles of O2. 

(39.09)+(35.45)+(3*15.99)= 122.51 g/ mol = molar mass of KClO3 
45.8 g KClO3/ 122.51 g/ mol KClO3 = .374 moles KClO3 
.374 mol KClO3 *(3 moles O2/2 mol KClO3)= .560 moles O2 

Once we have moles of O2, we can convert to grams of O2. 

(2*15.99)= 31.98 g/mol = molar mass of O2 
(.560 moles O2) (31.98 g/mol)= 17.91 g O2 

Hope this helps :)

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When 1 mol of a fuel burns at constant pressure, it exchanges-3452 kJ of heat and does-11 kJ of workon the surroundings. What ar

Degger [83]

Answer:-3463 kJ and -3452kJ

Explanation:

ΔU is the change in internal energy of a system and its formula is;

ΔU = q + w

Where q represents heat transferred into or out of the system. Its value is positive when heat is transfer into the system and negative when heat is produced by the system.

W represents the work done on or by the system. Its value is positive when work is done on the system and negative when it is done by the system.

For the system in this question, we see that it produces heat which means heat is transferred out of the system, therefore the value of q is negative, it can also be seen that work is done by the system which means that w is also negative.

Therefore,

ΔU = -q-w

ΔU = -3452 kJ – 11kJ

= - 3463kJ

ΔH is the change in the enthalpy of a system and its formuls is;

ΔH = ΔU + Δ(PV)

By product rule Δ(PV) becomes ΔPV + PΔV

At constant pressure ΔP = 0. Therefore,

ΔH = -q-w + PΔV

w is equals to PΔV, So:

ΔH = -q

ΔH = -3452kJ

6 0

2 years ago

In the winter, a heated home in the Northeast might be maintained at a temperature of 78°F. What is this temperature on the Cels

masya89 [10]

Answer: $25.6^0C$ , $298.6K$

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like $^0C$ and $K$

These units of temperature are inter convertible.

We are given:

Temperature of the gas = $78^0F$

Converting this unit of temperature into $^0C$ by using conversion factor:

$^oC=\frac{5}{9}\times (^oF-32)$

$^0C=\frac{5}{9}\times (78^oF-32)$

$25.6^0C$

Converting this unit of temperature into $K$ by using conversion factor:

$K=t^0C+273$

$K=25.6+273$

$298.6K$

Thus the temperature on the Celsius and Kelvin scales are $25.6^0C$ and $298.6K$ respectively.

6 0

2 years ago

The first-order reaction, so2cl2 → so2 + cl2, has a rate constant equal to 2.20 × 10-5 s-1 at 593 k. what percentage of the init

Svetradugi [14.3K]

For the 1st order reactions,rate constant (k) is mathematically expressed as

k = $\frac{2.303}{t}log \frac{Co}{Ct}$
where, t = time
Co = initial conc. of reactant
Ct = conc. of reactant after time 't'

Given: k = 2.20 × 10^-5 s-1, t = 2 hours = 7200 s

Therefore, we have
2.20 × 10^-5 = $\frac{2.303}{7200}log \frac{100}{Ct}$
∴ $log\frac{100}{Ct}$ = 0.06877
∴, $\frac{100}{Ct}$ = 1.1716
∴, Ct = 85.35%

Thus, 85.35 % of the initial amount of SO2Cl2 will remain after 2.00 hours.

4 0

2 years ago

What represents the hemisphere

Natasha_Volkova [10]

A line. You can even check a map

6 0

2 years ago

As more bulbs are connected in series in a circuit, what happens to the brightness with which each glows?

madam [21]

They dim unless more volts are added (batteries")

8 0

2 years ago

Read 2 more answers