Twh-bmyd-aer please j.o.i.n...

Uniones que se dan entre atomo y moleculas

lesya [120]

Answer:

khbfd skjb skbusf szkbs

Explanation:jdfh sms kok

6 0

3 years ago

Write electron configurations for Gallium, Ga (Z=31), and show the total valence electrons

ivann1987 [24]

Answer: The electronic configuration of gallium is written below and number of valence electrons is 3.

Explanation:

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

Valence electrons are defined as the electrons present in the outermost shell of an atom.

We are given:

An element Gallium having atomic number as 31.

Number of electrons = 31

Electronic configuration of Gallium is: $(Z=31):1s^22s^22p^63s^23p^64s^23d^{10}4p^1$

This element has 3 electrons in its outermost shell. So, the number of valence electrons is 3

Hence, the electronic configuration of gallium is written below and number of valence electrons is 3.

3 0

4 years ago

List the official SI units

Juli2301 [7.4K]

Answer:

The seven SI base units, which are comprised of:

Length - meter (m)

Time - second (s)

Amount of substance - mole (mole)

Electric current - ampere (A)

Temperature - kelvin (K)

Luminous intensity - candela (cd)

Mass - kilogram (kg)

Explanation:

7 0

4 years ago

Read 2 more answers

Hello can you please help me to solve above fill in the blank questions

LiRa [457]

Answer:

C2H5OH is a formula of ethyl alchol or ethanol.

alkenes are characterisized by carbon-carbon double bond.

burning is an example of exothermic reaction

atomic number of element is x the symbol of its ion is +1 bcz it will lose one electron

if temperature increases then rate of chemical reaction increases

Explanation:

i hope this will help you :)

7 0

4 years ago

Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be

amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

5 0

3 years ago