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lbvjy [14]
3 years ago
14

Write a nuclear equation for the beta decay of Promethium-165

Chemistry
1 answer:
padilas [110]3 years ago
4 0

Answer:

see your answer in pic

Explanation:

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The VSEPR model was developed before any xenon compounds had been prepared. Thus, these compounds provided an excellent test of
Natalka [10]

The maximum amount of XeF4 that could be produced is 0.5 moles.

XeF4 = Xe (g) 2 F2 (g) (g)

Xe and F2 have a mole ratio of 1:2. Because of this, the reaction would be limited by F2 when there is 1 mole of Xe and 1 mole of F2.

<h3>What is mole ratio?</h3>

The mole ratio is the ratio of any two compounds' mole amounts that are present in a balanced chemical reaction.

A comparison of the ratios of the molecules required to accomplish the reaction is given by the balancing chemical equation.

A mole ratio is a conversion factor used in chemical reactions to link the mole quantities of any two compounds. A conversion factor's numbers are derived from the balanced chemical equation's coefficients.

To learn more about mole ratio from the given link:

brainly.com/question/14425689

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6 0
2 years ago
Two major discoveries of the gold foil experiment
tiny-mole [99]
Rutherford's gold foil experiment proved that there was a small, dense, positively charged nucleus at the center, which contained most of the mass of the atom. Which contained electrons orbiting the nucleus.
5 0
3 years ago
Which of the following lists elements from lowest to highest atomic number
vitfil [10]
Lowest is Hydrogen highest is <span>Beryllium
-HOPE THIS HELPED </span>
8 0
3 years ago
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 is the volume of 1.60 grams of O2 gas at STP? (5 points)
tiny-mole [99]

Answer:

  • <u><em>1.12 liters</em></u>

Explanation:

<u>Calculating number of moles</u>

  • Molar mass of O₂ = 32 g
  • n = Given weight / Molar mass
  • n = 1.6/32
  • n = 0.05 moles

<u>At STP</u>

  • One mole of O₂ occupies 22.4 L
  • Therefore, 0.05 moles will occupy :
  • 22.4 L x 0.05 = <u><em>1.12 L</em></u>
5 0
2 years ago
Read 2 more answers
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