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Gnoma [55]
3 years ago
10

Do you think a battery system would OR wouldn’t have energy? If so, give me evidence/observations as to why you think so.

Chemistry
1 answer:
Basile [38]3 years ago
6 0

Answer:

Yes it would

Explanation: Well it kinda depend on the voltage and how the battery has been in use or based on the condition

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A 10-liter container has 2 moles of oxygen at a pressure of 92 kpa. The effective speed (rms) of the oxygen molecules in the gas
bezimeni [28]

The effective speed (rms) of the oxygen gas is 293.68 m/s.

<h3></h3><h3>What is Root-mean-square velocity?</h3>

Root mean square velocity is the square root of the mean of squares of the velocity of individual gas molecules

v_{rms}=\sqrt[]{\frac{3RT}{M} }

<em>where </em>R = universal gas constant

M = molar mass of the gas in kg/mol

T = temperature in Kelvin

According to the ideal gas law,

PV = nRT

RT = \frac{PV}{n}

Substitute in the rms velocity formula,

v_{rms} = \sqrt[]{\frac{3PV}{nM} }

P = 92 kPa, V = 10 L, n = 2 moles and M = 32 x 10⁻³ kg/mol

v_{rms} = \sqrt[]{\frac{3\times92\times10}{2\times32\times10^-^3} }

=293.68 m/s

Thus, the effective speed (rms) of O₂ gas is 293.68 m/s.

Learn more about Root-mean-square velocity:

brainly.com/question/15995507

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4 0
1 year ago
Which of the following statements about protons are true?
attashe74 [19]
The second on it true
4 0
3 years ago
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Alguien me da la fórmula de : Fosfuro ácido de estroncio?
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la formula es Sr(OH)^2
6 0
3 years ago
The outside temperature is<br> 35°C. What is the temperature in<br> K
Fittoniya [83]

Answer:

308.15

Explanation:

308.15K − 273.15 = 35°C

5 0
3 years ago
A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g
Lynna [10]

Answer:

\%NaHCO_3=61.2\%

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

NaHCO_3+HA\rightarrow NaA+H_2O+CO_2

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%

Best regards.

5 0
3 years ago
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