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Firlakuza [10]
3 years ago
8

If the input force applied to one end of a first-class lever is 8 N, and the fulcrum is located 6 m away from the input force bu

t 3 m away from the output force, what will be the resulting output force?
Chemistry
1 answer:
MatroZZZ [7]3 years ago
4 0

Answer:

16 N

Explanation:

The ratio of output force to the input force is called mechanical advantage of the lever. Also, the ratio of input arm distance to the output arm distance is called mechanical advantage of the lever.

We have,

Input force = 8 N

Input arm distance = 6 m

Output arm distance = 3 m

We need to find the resulting output force. So,

\dfrac{F_o}{8}=\dfrac{6}{3}\\\\F_o=16\ N

So, the resulting output force is 16 N.

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It's 32°F (0°C) outside under normal atmospheric conditions (1 atm) at a stunt
Korolek [52]

The volume of the balloon is approximately 2652 liters.

<h3>How to determine the volume occupied by the gas in a balloon </h3>

Let suppose that <em>flammable</em> hydrogen behaves ideally. GIven the molar mass (M), in kilograms per kilomole, and mass of the gas (m), in kilograms. The volume occupied by the gas (V), in cubic centimeters, is found by the equation of state for <em>ideal</em> gases:

V = \frac{m\cdot R_{u}\cdot T}{P\cdot M}   (1)

Where:

  • R_{u} - Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin.
  • T - Temperature, in Kelvin
  • P - Pressure, in kilopascals

If we know that m = 0.239\,kg, R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}, T = 273.15\,K, P = 101.325\,kPa and M = 2.02\,\frac{kg}{kmol}, then the volume of the balloon is:

V = \frac{(0.239\,kg)\cdot \left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot (273.15\,K)}{\left(101.325\,kPa\right)\cdot \left(2.02\,\frac{kg}{kmol} \right)}

V = 2.652\,m^{3} (2652\,L)

The volume of the balloon is approximately 2652 liters. \blacksquare

To learn more on ideal gases, we kindly invite to check this verified question: brainly.com/question/8711877

4 0
2 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
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Atomic mass is the answer 

if u hav any  more questions in chemistry ask me! I am very good at chemistry.
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22 Newton’s to the right but I’m not to sure make sure to check other answers!
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Electron -  In an atom, the central part which is composed of neutrons and protons including quarks is called the nucleus. Whereas the electron is the particle with negative elementary electric charge can be found inside the atom surrounding the nucleus. 
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