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3 years ago

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What is ionization energy

1 answer:

damaskus [11]3 years ago

5 0

Ionization energy is the energy required to remove an electron from a gaseous atom or ion. The first or initial ionization energy or Ei of an atom or molecule is the energy required to remove one mole of electrons from one mole of isolated gaseous atoms or ions

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A 40.0-mL sample of 0.100 M HNO2 (Ka = 4.6 x 10-4 .) is titrated with 0.200 M KOH. Calculate: a. the pH when no base is added b.

In-s [12.5K]

Answer:

a) 2.173

b) 20ml

c) 3.87

d) 4.35

e) 8.8

f) 12.45

Explanation:

chech the attachment for explanationns

4 0

2 years ago

An acid is any substance that

ehidna [41]

Answer:

C

Explanation: a is incorrect since the lower the ph = more acidic and b is incorrect because it produces hydronium ion and d I’m not sure what it is but I no that base recieve the protons

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2 years ago

It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry

vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

$\frac{dP}{dT} = \frac{L}{T \times \Delta V}$

Also, $\frac{L}{T} = \Delta S^{fusion}_{m}$

where, $\Delta V^{fusion}_{m}$ is the difference in specific volumes.

Hence, $\frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}$

As, $\Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15}$ = 22.0 J/mol K

And, $\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$ ...... (1)

where, $d_{H_{2}O}$ = density of water

$d_{ice}$ = density of ice

M = molar mass of water = $18.02 \times 10^{-3} kg$

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

$\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$

= $\frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}$

= $-1.51 \times 10^{-6}$

Therefore, calculate the required pressure as follows.

$\frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}$

= $1.45 \times 10^{7} Pa/K$

or, = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP = $145 \times 4.7 bar$

= 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

5 0

2 years ago

Since the electronegativity of O > C = H, the shared electrons in the covalent bonds of carbon dioxide (CO2) and methane (CH4

Stells [14]

Answer: option B. closer to oxygen in the case of CO2

Explanation:

5 0

2 years ago

Fill in the blanks.

vichka [17]

Answer:

Hello There!!

Explanation:

1.C nH 2n

2.Family

3.carbonyl group

4.Catalyst

5.Nickel

hope this helps,have a great day!!

~Pinky~

4 0

2 years ago

Read 2 more answers