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zlopas [31]
2 years ago
7

A 1.0 L volume of gas at 27.0°C exerts a pressure of 85.5 K PA what will the pressure be at 127°C assume constant volume

Chemistry
1 answer:
Marianna [84]2 years ago
4 0

Answer:

114 kPa

Explanation:

Using Gay-Lussac's law you get the equation \frac{P1}{T1} x \frac{P2}{T2} and converting celcius you get the final equation of \frac{85.5}{27+273}  x \frac{P2}{127+273} . After dividing 85.5 by 27+273(300) you get 0.285 and then you multiply 0.285 by 127+273 (400). You finally get 114 kPa

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A, O2 has to be a reactant for combustion to burn
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3 years ago
Wuy Piolcululugu<br> Give the names of two organs in the chest.<br> 1. ...............<br> 2 m
Arada [10]

Answer:

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1. The lungs

2. The heart

Explanation:

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This contain the lungs , the tracheobronchial tree , the heart , the blood vessels which transport the blood between the heart and the lungs.

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3 years ago
4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a
yawa3891 [41]

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

6 0
3 years ago
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stiv31 [10]

Answer:

3.90*10^{-25}kg/atom

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The molar mass of uranium-235 is 235 g/mol. So one mole of uranium-235 has a mass of 235 g. Put differently 6.022×10^23 atoms of uranium-235 have a mass of 235 g. Knowing that, how can we use that to find the mass of one atom?

mass of one atom = \frac{235 g}{1mol} *\frac{1 mol}{6.022*10^{23}atoms } *\frac{1kg}{1000g}= 3.90*10^{-25}kg/atom

7 0
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anygoal [31]
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7 0
3 years ago
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