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4 years ago

How many protons and electrons are present in Au+?

Chemistry

1 answer:

balandron [24]4 years ago

7 0

Answer:

Protons: 79

Electrons: 78

Explanation:

1. The number of protons is the atomic number (The atomic number for Au on the periodic table is 79)

2. Since the charge is +1 (positive) it means that there's one more proton than electrons. So, 79-1 = 78 electrons

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Use de Broglie's hypothesis to determine the speed of the electron in a hydrogen atom when in the n = 5 orbit, whose radius is 1

andriy [413]

Answer:

Velocity = 4.41×10⁵ m/s

Explanation:

The force of attraction between the nucleus and the electron is equal to the centrifugal force acting on the outermost electron due to circular rotation. It can be mathematically written as:

$\frac {K\times q^2}{r^2}=\frac {m\times v^2}{r}$

Where,

K is the Coulomb's constant having value 9×10⁹ N. m²/C²

q is the charge on the electron or the proton which is 1.6×10⁻¹⁹ C

r is the radius of the atom

m is the mass of the electron having value 9.1×10⁻³¹ Kg

v is the velocity of the electron.

The equation can be written to calculate the velocity as:

$v=\sqrt {\frac {K\times q^2}{m\times r}}$

Given that:

r = 1.3×10⁻⁹ m

So,

$v=\sqrt {\frac {9\times 10^9\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times 1.3\times 10^{-9}}}$

<u>Velocity = 4.41×10⁵ m/s</u>

4 0

3 years ago

What can affect the properties of a substance?

JulsSmile [24]

The answer is D: only the arrangement of atoms.

6 0

3 years ago

If helium effuses through a porous barrier in 1.34 min, how much time (in min) would it take the same amount of ammonia to effus

timofeeve [1]

Answer:

The time taken the same amount of ammonia to effuse through the same barrier under the same conditions is 2.76 minutes.

Explanation:

Let the volume of the helium gas be = V

Time taken by the helium gas = t = 1.34 min

Effusion rate of helium gas = $R=\frac{V}{1.34 min}$

If V volume of ammonia effuse through same porous barrier the effusion rate of ammonia gas will be given as:

$R'=\frac{V}{t'}$

Using Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

Molar mass of helium gas = M = 4 g/mol

Molar mass of ammonia gas = M' = 17 g/mol

$\frac{R}{R'}=\sqrt{\frac{M'}{M}}$

$\frac{\frac{V}{1.34 min}}{\frac{V}{t'}}=\sqrt{\frac{17 g/mol}{4 g/mol}}$

$t'=1.34 min\times \sqrt{\frac{17 g/mol}{4 g/mol}}=2.76 min$

The time taken the same amount of ammonia to effuse through the same barrier under the same conditions is 2.76 minutes.

8 0

3 years ago

Consider the following reaction where Kp = 2.01 at 500 K: PCl3(g) + Cl2(g) PCl5(g) If the three gases are mixed in a rigid conta

alekssr [168]

Explanation:

Kp remains constant (if T=const.).

If Q<Kp, more reactants are consumed (the direct reaction is in progress). If Q>Kp the reverse reaction is in progress (the products are consumed).

1. A reaction will occur in which PCl5 (g) is consumed

A. If Kp > Qp then 1,2,3 and 5 are F. 4 is T

B. Kc > Qc then 1,3 are T. 2,4,5 are F.

C. Qp > Kp then 1.3.4 are T. 2,5 are F.

7 0

4 years ago

Two isotopes of lithium are found in nature Li6 has a mass of 6. 02u and Li7 has a mass of 7.02u . Use the atomic weight of lith

alexandr402 [8]

The isotope that is more abundant, given the data is isotope Li7

<h3>Assumption</h3>

Let Li6 be isotope A
Let Li7 be isotope B

<h3>How to determine whiche isotope is more abundant</h3>

Molar mass of isotope A (Li6) = 6.02 u
Molar mass of isotope B (Li7) = 7.02 u
Atomic mass of lithium = 6.94 u
Abundance of A = A%
Abundance of B = (100 - A)%

Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]

6.94 = [6.02A% + 702 - 7.02A%] / 100

Cross multiply

6.02A% + 702 - 7.02A% = 6.94 × 100

6.02A% + 702 - 7.02A% = 694

Collect like terms

6.02A% - 7.02A% = 694 - 702

-A% = -8

A% = 8%

Thus,

Abundance of B = (100 - A)%

Abundance of B = (100 - 8)%

Abundance of B = 92%

SUMMARY

Abundance of A (Li6) = 8%
Abundance of B (Li7) = 92%

From the above, isotope Li7 is more abundant.

Learn more about isotope:

brainly.com/question/24311846

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4 0

2 years ago