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gladu [14]
2 years ago
7

I need help pleaseee!!!!!!!! Fast

Chemistry
2 answers:
VikaD [51]2 years ago
6 0

Answer:

C.) 1700 Pa

Explanation:

300 k - 0.3 m^3 = 17000

AfilCa [17]2 years ago
4 0

Answer:

C is the correct answer

Explanation:

Hope this helps:)

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denis-greek [22]
Yes a nucleus is located inside a cell.
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Please hurry i’m taking a test
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10 N to the right

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What mass of ch4 is consumed when 587. 9 kj of energy is evolved from the combustion of a mixture of ch4(g) and o2(g)? show the
Juliette [100K]

The mass of CH₄ that is consumed is 10.598 g.

<h3>What is enthalpy?</h3>

The enthalpy of a system is defined as the sum of the internal energy of a system and the energy that is produced due to its pressure and volume.

It is given by

H = U+PV

The following reaction takes place during the combustion of CH₄

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

Energy evolved during the combustion, E = 587.9 kJ

Enthalpy of combustion of methane, ∆H = 890.3 kJ

No of moles of CH_4 =\frac{energy evolved}{enthalpy of combustion}

= 587.9 /890

Mass of CH₄  that is consumed = no of moles × molar mass of CH₄

  = 587.9 /890 × 16

   = 10.598 g

The mass of CH₄  which is consumed is 10.598 g

Learn more about enthalpy:

brainly.com/question/14047927

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6 0
1 year ago
What occurs when a coordinate covalent bond is formed between nitrogen and hydrogen in the ammonium ion, NH4+?
NeX [460]
Covalent bonds are formed when electrons are shared between elements that are nonmetals. The ammonium ion, NH+4 , would have covalent bonds because both nitrogen and hydrogen are nonmetals. ... So, the bond between this particular hydrogen atom and the central nitrogen is a dative covalent bond.
4 0
2 years ago
Determine the equilibrium constant, Keq, at 25°C for the reaction
adelina 88 [10]

Explanation:

The given chemical reaction is:

2Br^- (aq) + I_2(s)  Br_2(l) + 2I^- (aq)

E^ocell=oxidation potential of anode + reduction potential of cathode\\

The relation between Eo cell and Keq is shown below:

deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

E^ocell= (-1.07+0.53)V\\=-0.54V

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3

Answer:

Keq=6.13x10^33

3 0
2 years ago
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