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3 years ago

A gold atom in its neutral state has how many electrons

1 answer:

riadik2000 [5.3K]3 years ago

4 0

To find the number of electrons in a neutral gold atom look at it's atomic number. The atomic number is the number of protons in the element. For an atom to be neutral it needs the same number of electrons as protons, therefore the positive and negative charges will be balanced out. Gold's atomic number is 79 therefore it has 79 electrons.

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In what situations are van der waals forces the most significant

viktelen [127]

Its most significant when all other forces are absent

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3 years ago

Why is a thermocouple a good device for indicating the temperature<br><br> in a car engine?

Anna35 [415]

A thermocouple is a sensor used to measure temperature. Thermocouples are made with two wires of different metals, joined together at one end to form a junction. ... Naturally, a thermocouple outputs a millivolt signal, therefore, as the resistance changes, the change in voltage can be measured.

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3 years ago

Read 2 more answers

What is the oxidation number of cl− in the hypochlorite ion clo−?

mariarad [96]

Answer : The oxidation number of chlorine (Cl) is, (+1)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

The given compound is, $ClO^-$

Let the oxidation state of 'Cl' be, 'x'

$x+(-2)=-1\\\\x-2=-1\\\\x=+1$

Therefore, the oxidation number of chlorine (Cl) is, (+1)

5 0

3 years ago

What volume is occupied by 0.109 molmol of helium gas at a pressure of 0.98 atmatm and a temperature of 307 K

KATRIN_1 [288]

Answer:

2.8 L

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 0.109 mole

Pressure (P) = 0.98 atm

Temperature (T) = 307 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

The volume of the helium gas can be obtained by using the ideal gas equation as follow:

PV = nRT

0.98 × V = 0.109 × 0.0821 × 307

0.98 × V = 2.7473123

Divide both side by 0.98

V = 2.7473123 / 0.98

V = 2.8 L

Thus, the volume of the helium gas is 2.8 L.

6 0

2 years ago

When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C

natka813 [3]

Answer:

$\Delta_{r}U$ of the reaction is -6313 kJ/mol

$\Delta_{r}H$ of the reaction is -6312 kJ/mol

Explanation:

$Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C$

$q_{cal}= C \times \Delta T$

$=5.861 \times 3.596 = 21.076\,kJ$

$q_{rxn}= -q_{cal}= -21.076\,kJ$

$\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol$

Therefore, $\Delta_{r}U$ of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter is as follows.

$C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)$

$Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5$

$\Delta_{r}H=\Delta E+ \Delta n. RT$

$=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol$

Therefore, $\Delta_{r}H$ of the reaction is -6312 kJ/mol.

3 0

3 years ago