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myrzilka [38]
3 years ago
5

calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2​

Physics
1 answer:
igor_vitrenko [27]3 years ago
7 0

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

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A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina
Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy U=4.30\times10^{3}\ J

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

\Delta S=\dfrac{\Delta U}{T}

Where, \Delta U = internal energy

T = average temperature

Put the value in to the formula

\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}

\Delta S=-14.72\ J/K

Hence, The approximate change in entropy is -14.72 J/K.

5 0
3 years ago
A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0
3 years ago
Calculate the potential energy of a 5 kg object sitting at the top of a 2 meter ramp.
trapecia [35]
The formula used to find potential energy is <em>P.E. = M * G * H</em> (P.E. is potential energy, M is mass, G is gravitational pull, and H is height). So the answer to your question is <em>5 * 9.8 * 2</em>, which equals 98.
5 0
3 years ago
Energy caused by the flow of ocean water
Arlecino [84]
It’s thermal energy
8 0
2 years ago
Block A of 5 kg with a speed of 3 m/s collides with block B of 10 kg that is stationary. After the collision, block B travels wi
Lemur [1.5K]

Answer:

-1m/s

Explanation:

We can calculate the speed of block A after collision

According to collision theory:

MaVa+MbVb = MaVa+MbVb (after collision)

Substitute the given values

5(3)+10(0) = 5Va+10(2)

15+0 = 5Va + 20

5Va = 15-20

5Va = -5

Va = -5/5

Va = -1m/s

Hence the velocity of ball A after collision is -1m/s

Note that the velocity of block B is zero before collision since it is stationary

6 0
3 years ago
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