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viktelen [127]
3 years ago
9

If the mass of an object is 10 kg and the acceleration of an object due to

Chemistry
1 answer:
mars1129 [50]3 years ago
5 0
Hmm the answer is 2.454

No explanation needed it’s just 2.454
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This is an impossible formula. Pretend it is real.
guapka [62]

There are eight moles of oxygen atoms  in 1 mole of Mn_{9}(ClO_{4}) _{2}.

<h3>What is the number of moles of oxygen atoms?</h3>

We know that a compound is composed of atoms. The atoms that make up the molecule are chemically combined. It is usual that the number of atoms in the compound would correspond with the chemical formula.

Now we have the compound  Mn_{9}(ClO_{4}) _{2}. In one mole of the compound we have;

  • 9 Moles of manganese atom
  • 2 moles of chlorine atom
  • 8 moles of oxygen atom

Thus, there are eight moles of oxygen atoms  in 1 mole of Mn_{9}(ClO_{4}) _{2}.

Learn more about atoms;brainly.com/question/1566330

#SPJ1

7 0
2 years ago
What are the three states of matter, describe atoms during these states.
Dominik [7]

Answer:

Three states of matter is liquid, solid and gas.

I haven't learned atoms, I just learned the states in 5th grade.

7 0
3 years ago
Read 2 more answers
∆G° for the reaction of nitrogen with hydrogen to produce ammonia (see balanced chemical equation below) has a value of -33.3 kJ
Juliette [100K]

Answer:

-0.85KJ

Explanation:

Given N2(g) + H2(g) <--->2NH3(g)

Kp =[ P(NH3)]²/[P(H2)]³[P(N2)]

Where P is the pressure of the gas

P(H2)b= P(N2) = 125atm

P(NH3) = 200atm

Kp = 2²/(125)³(125)

Kp = 2.048 ×10^-6

∆G = -RTlnKp

R =0.008314 J/Kmol

T = 25 +273/= 298k

= 8.314 ×10^-3 × 298 × ln(2.048 ×10^-6)

= -0.008314 × 298 × (-13.099)

= 32.45KJ

∆G = ∆G° + RTlnKp

∆G = -33.3 + 32.45

∆G = -0.85KJ or -850J

3 0
4 years ago
Lonization energy is the
Tamiku [17]
Arrow on the table below draw an
3 0
3 years ago
The total volume of seawater is 1.5 x 10²¹ L. Seawater contains approximately 3.5% sodium chloride by mass. At that high of a co
garri49 [273]

Answer:

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater (m_{sw}), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

m_{sw} = \rho_{sw}\cdot V_{sw}

Where:

\rho_{sw} - Density of seawater, measured in grams per liters.

V_{sw} - Volume of seawater, measured in liters.

If V_{sw} = 1.5\times 10^{21}\,L and \rho_{sw} = 1030\,\frac{g}{L}, then:

m_{sw}=\left(1030\,\frac{g}{L} \right)\cdot (1.5\times 10^{21}\,L)

m_{sw} = 1.545\times 10^{24}\,g

The total mass of sodium chloride is determined by the following ratio:

r = \frac{m_{NaCl}}{m_{sw}}

m_{NaCl} = r\cdot m_{sw}

Given that m_{sw} = 1.545\times 10^{24}\,g and r = 0.035, the total mass of sodium chloride in all the seawater in the world is:

m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)

m_{NaCl} = 5.408\times 10^{22}\,g

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

8 0
4 years ago
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