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Rasek [7]
3 years ago
10

Why is freshly distilled or deionized water used in this standardization?

Physics
2 answers:
sukhopar [10]3 years ago
8 0

Answer:

The amount of carbon dioxide is little in deionized water.

Explanation:

Deionized water is a water with little or no impurities. Impurities are in waters are not able to boil below or above the boiling point of water,and in this case are been retained in the original container.

erastova [34]3 years ago
5 0

Answer:

There is little CO2 in freshly deionized or distilled water.

Explanation:

Freshly distilled water has no hydrogen peroxide. Distilled water contains calcium, which stabilizes the sodium hydroxide , Freshly deionized water has a pH of 4.72. There is little CO2 in freshly deionized or distilled water.

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What is the atmospheric pressure and temperature at sea level in a standard<br> atmosphere?
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Answer:

The tropospheric tabulation continues to 11,000 meters (36,089 ft), where the temperature has fallen to −56.5 °C (−69.7 °F), the pressure to 22,632 pascals (3.2825 psi), and the density to 0.3639 kilograms per cubic meter (0.02272 lb/cu ft). Between 11 km and 20 km, the temperature remains constant

Explanation:

Hope this helped, Have a wonderful day!!

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Electrons in an atom make up most of the _________ of the atom, while the protons and neutrons make up nearly all of the _______
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Which of the following does not change the resistance of a wire?
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An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elec
vaieri [72.5K]

Answer:

Speed = 575 m/s

Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.

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Potential difference, U = $-3.45 \times 10^{-3} \ V$

Mass of the alpha particle, $m_{\alpha} = 6.68 \times 10^{-27} \ kg$

Charge of the alpha particle is, $q_{\alpha} = 3.20 \times 10^{-19} \ C$

So the potential difference for the alpha particle when it is accelerated through the potential difference is

$U=\Delta Vq_{\alpha}$

And the kinetic energy gained by the alpha particle is

$K.E. =\frac{1}{2}m_{\alpha}v_{\alpha}^2 $

From the law of conservation of energy, we get

$K.E. = U$

$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$

$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$

$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$

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