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nata0808 [166]
3 years ago
12

What is au electron configuration

Chemistry
1 answer:
ozzi3 years ago
6 0
Au = 79

1s22s22p63s23p63d104s24p64d105s25p64f145d106s1

Or

[Xe] 4f14 5d10 6s1
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when we drink the milk the digestive system produce protease enzyme to break down the protein of milk.

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In this experiment, you will measure all masses in
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0.498 kg

Explanation:

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Which term defines the distance from the focal point to either a lens or a mirror
nikklg [1K]
The focal length, like you said it's the distance between the FOCAL point and the mirror.


Hope this helps.... :)
3 0
2 years ago
A solution was made by dissolving 5.10 mg of hemoglobin in water to give a final volume of 1.00 ml. the osmotic pressure of this
Assoli18 [71]

The molecular weight of hemoglobin can be calculated using osmotic pressure

Osmotic pressure is a colligative property and it depends on molarity as

πV = nRT

where

π = osmotic pressure

V = volume = 1mL = 0.001 L

n = moles

R = gas constant = 0.0821 L atm / mol K

T = temperature = 25°C = 25 + 273 K = 298 K

Putting values we will get value of moles

moles=\frac{\pi V}{RT}=\frac{0.00195X0.001}{0.0821X298}mol

we know that

moles=\frac{mass}{molarmass}

Therefore

molarmass=\frac{mass}{moles}=\frac{5.10X10^{-3}g}{7.97X10^{-8}}=6.399X10^{4}g

5 0
2 years ago
If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera
Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0
2 years ago
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