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iragen [17]
3 years ago
7

WILL GIVE BRAINLY

Chemistry
1 answer:
cestrela7 [59]3 years ago
8 0
KCIO3 → KCI + O2

First you have to count all the atoms of each element in both sides of the reaction, in this case:
Left side: there is one K atom, one Cl atom and 3 O atoms
Right side: there is one K atom, one Cl atom and 2 O atoms

K and Cl are balanced but O isn't
To balance O you can add a coefficient to O2 in the right side of the equation like this:

KCIO3 → KCI +3 O2

Now you have 6 oxygen atoms in the right side but only 3 in the left side, so you can add a coefficient to KClO3:

2KCIO3 → KCI +3 O2

And finally, Balance K and Cl again by adding a 2 as a coefficient for KCl

2KCIO3 → 2KCI +3 O2

Now both sides have the same amount of atoms of each element (2 K atoms, 2 Cl atoms and 6 O atoms)

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Lelu [443]
You would get 35.................
4 0
2 years ago
34 grams of carbon react with an unlimited amount of H2O. The reaction is: C + H2O → CO + H2 What is the ending substance?
katovenus [111]
H2c+o
hope this helps :0

7 0
2 years ago
Read 2 more answers
Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

6 0
3 years ago
What volume of oxygen (in L) is produced
sveticcg [70]

Answer:

12.36 L.

Explanation:

We'll begin by calculating the number of mole in 147.1 g of lead(II) nitrate, Pb(NO₃)₂. This can be obtained as follow:

Molar mass of Pb(NO₃)₂ = 207.2 + 2[14.01 + (16×3)]

= 207.2 + 2[14.01 + 48]

= 207.2 + 2[62.01]

= 207.2 + 124.02

= 331.22 g/mol

Mass of Pb(NO₃)₂ = 147.1 g

Mole of Pb(NO₃)₂ =?

Mole = mass / Molar mass

Mole of Pb(NO₃)₂ = 147.1 / 331.22

Mole of Pb(NO₃)₂ = 1.104 moles.

Next, we shall determine the number of mole of oxygen gas, O₂, produce from the reaction. This can be obtained as follow:

2Pb(NO₃)₂ —> 2PbO + 4NO₂ + O₂

From the balanced equation above,

2 moles of Pb(NO₃)₂ decomposed to produce 1 mole of O₂.

Therefore, 1.104 moles of Pb(NO₃)₂ will decompose to produce = (1.104 × 1)/2 = 0.552 mole of O₂.

Finally, we shall determine the volume occupied by 0.552 mole of oxygen gas, O₂. This can be obtained as follow:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 0.552 mole of O₂ will occupy = 0.552 × 22.4 = 12.36 L at STP.

Thus, the volume of oxygen gas, O₂ produced is 12.36 L.

6 0
2 years ago
What product has almost the same compounds as baking soda 10pts
mamaluj [8]
Ummm i actually dont know lol i swear i knew im sorry man or women ( im not trying to assume your gender 
4 0
3 years ago
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