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ELEN [110]
3 years ago
5

Calculate the distance Jupiter in miles if it has an AU of 5.2

Chemistry
1 answer:
svetlana [45]3 years ago
3 0
About 484 million miles
Fifth Planet from Our Star
Jupiter orbits about 484 million miles (778 million kilometers) or 5.2 Astronomical Units (AU) from our Sun (Earth is one AU from the Sun).
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How many protons electrons and neutrons does the isotope nitrogen 15 have
Ivenika [448]
Answer:

The atomic number of<span>N<span>157</span></span> 
The number of protons is 7
The number of electrons is 7
The number of neutrons is 8

Explanation:

The atomic number of Nitrogen is 7 because Nitrogen has 7 protons. 
The seven protons attract 7 electrons in the ground state.

If the atom had fewer or more than 7 protons the atom would not be Nitrogen.

The mass of the atom is the sum of protons and neutron. so

p + n = mass ( protons (p) and neutrons(n) both have an atomic mass of one

7 + n = 15 subtract 7 from both sides

<span>7−7+n=15−7</span>

n = 8

4 0
3 years ago
In which of the following are the symbol and name for the ion given correctly
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3 years ago
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All of the following are important properties of prokaryotes except A. prokaryotes are often decomposers and enable recycling of
Sladkaya [172]
The answer to this is B.
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3 years ago
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At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initiall
mestny [16]

Answer:

concentration of [O_2] = 0.0124 = 12.4 ×10⁻³ M

concentration of [CO] = 0.0248 = 2.48 ×10⁻² M

concentration of [CO_2] = 0.4442 M

Explanation:

Equation for the reaction:

2CO_2_{(g)                ⇄          2CO_{(g)       +       O_2_{(g)

Concentration of   CO_2_{(g) = \frac{2.3}{4.9}  = 0.469

For our ICE Table; we have:

                       2CO_2_{(g)                ⇄          2CO_{(g)       +       O_2_{(g)

Initial                 0.469                              0                           0

Change              - 2x                                +2x                      +x

Equilibrium       (0.469-2x)                       2x                         x

K = \frac{[CO]^2[O]}{[CO_2]^2}

K = \frac{[2x]^2[x]}{[0.469-2x]^2}

4.1*10^{-6}=\frac{2x^3}{(0.469-2x)^2}

Since the value pf K is very small, only little small of  reactant goes into product; so (0.469-2x)² = (0.469)²

4.1*10^{-6} = \frac{2x^3}{(0.938)}

2x^3 =3.8458*10^{-6

x^3 =\frac{3.8458*10^{-6}}{2}

x^3=1.9229*10^{-6

x=\sqrt[3]{1.9929*10^{-6}}

x = 0.0124

∴ at equilibrium; concentration of  [O_2] = 0.0124 = 12.4 ×10⁻³ M

concentration of [CO] = 2x  = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of [CO_2] = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M

3 0
3 years ago
What did the name carbon come from
Simora [160]
<span>The English name carbon comes from the Latin carbo for coal and charcoal.</span>
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3 years ago
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