Let's use the example: H2O ---> H2 + O2
We find how many elements of a product are on one side and how many elements on the other side.
Reactant: H=2 O=1
Product: H=2 O=2
We need to make the same amount of hydrogen and oxegyn atoms on each side, regardless of how high the numbers are, and we do this by adding coefficients to the compounds.

Reactant: H=4 O=2
Product : H=4 O=2
2 H2O---> 2 H2 + O2

8 0

2 years ago

The best determinant of how an individual measures their quality of life is _______.

Ronch [10]

C is the correct answer

4 0

3 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

Please help i am striggling with science so bad

Dennis_Churaev [7]

well, yes it will continue to swing but not forever. it will just a long time but eventually stop. the reason is because of the air resistance which will continue to damp the motion until the bob stops

3 0

2 years ago