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2 years ago

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Chemistry

1 answer:

jok3333 [9.3K]2 years ago

7 0

Answer:

1 mole represents 6.023×1023 particles.

1 mole of iodine atom= 6.023×1023

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg

Given 48g of Mg = 2×6.023×1023

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 1023

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 1023

no. of chlorine atom = 6.023×1023

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 1023

no. of atoms of hydrogen= 4 moles of hydrogen atom.

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SEP Analyze Data Look at the phase diagram. What is the state of matter for methanol at a temperature of −0.1°C and a pressure o

devlian [24]

If it’s a negative number and a positive it might be 1.4

8 0

2 years ago

An isotope of potassium with a half-life of roughly one billion years is often used in radioactive dating. This isotope decays t

melisa1 [442]

Answer : The chemical symbol for the element that results from this process is, (Ar) for argon.

Explanation :

Electron capture : In this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. Simply, a proton and an electron combines together to form a neutron. Mass number does not change in this process.

$_Z^A\textrm{X}+_{-1}^0\textrm{e}\rightarrow _{Z-1}^A\textrm{Y}$

The equation for the given reaction is,

$_{19}^{40}\textrm{K}+_{-1}^0\textrm{e}\rightarrow _{18}^{40}\textrm{Ar}$

Thus, the chemical symbol for the element that results from this process is, argon (Ar).

7 0

3 years ago

How many atoms are in 3.20 moles of carbon

DIA [1.3K]

Answer:

19.264× $10^{23}$ atoms are present in 3.2 moles of carbon.

Explanation:

It is known that one mole of each element is composed of Avagadro's number of atoms. This is same for all the elements in the periodic table.

So, as 1 mole of any element = Avagadro's number of atoms = 6.02× $10^{23}$ atoms

It is as simple as understanding a dozen of anything is equal to 12 pieces of that object.

As here the moles of carbon is given as 3.20 moles, the number of atoms in this mole can be determined as below.

1 mole of carbon = 6.02 × $10^{23}$ atoms

Then, 3.20 moles of carbon = 3.20 × 6.02 × $10^{23}$ atoms

Thus, 19.264× $10^{23}$ atoms are present in 3.2 moles of carbon.

6 0

2 years ago

PLEASE HELP & SHOW WORK

Olegator [25]

Answer:

1. 0.178 moles ; 2. 8x10²³ atoms ; 3. 7.22x10²³ molecules ; 4. 89.6 g ; 5. 1.34x10²² atoms ; 6. 1.67x10²⁵ molecules

Explanation:

1. Mass / Molar mass = Mol

5g / 28 g/m = 0.178 moles

2. 1 molecule of N₂ has 2 atoms, it is a dyatomic molecule.

4x10²³ x2 = 8x10²³ atoms

3. 1 mol of anything, has 6.02x10²³ particles

6.02x10²³ molecules . 1.2 mol = 7.22x10²³

4. 1 atom of C weighs 12 amu.

4.5x10²⁴ weigh ( 4.5x10²⁴ . 12) = 5.24x10²⁵ amu

1 amu = 1.66054x10⁻²⁴g

5.24x10²⁵ amu = (5.24x10²⁵ . 1.66054x10⁻²⁴) = 89.6 g

5. Molar mass NaCl = 58.45 g/m

1.3 g / 58.45 g/m = 0.0222 moles

1 mol has 6.02x10²³ atoms

0.0222 moles → ( 0.0222 . 6.02x10²³) = 1.34x10²²

6. Density of water is 1 g/mL, so 500 mL are contained in 500 g of water

Molar mass H₂O = 18 g/m

500 g / 18 g/m = 27.8 moles

6.02x10²³ molecules . 27.8 moles = 1.67x10²⁵

8 0

3 years ago

For the following electrochemical reaction: Al3+(aq) + 3e -> Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -> 2F (aq) Calculate

makvit [3.9K]

<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

$E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=2.87-(-1.66)=4.53V$

Hence, the standard electrode potential of the cell is 4.53 V.

6 0

2 years ago