Answer:
Anything that shares electrons.
Explanation: This means they don't transfer electrons.
Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.
Solution :
Using ideal gas equation,
where,
n = number of moles of gas
w = mass of ammonia gas = ?
P = pressure of the ammonia gas = 2.55 atm
T = temperature of the ammonia gas =
M = molar mass of ammonia gas = 17 g/mole
R = gas constant = 0.0821 L.atm/mole.K
V = volume of ammonia gas = 3.00 L
Now put all the given values in the above equation, we get the mass of ammonia gas.
Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.
Answer: GeH4 (Germanium(IV) Hydride)
Explanation:
A Binary molecular compound Hydrogen and a Group 4A element which is more more acidic than SiH4 in aqueous solution is GeH4.
The pKa of GeH4;
= 25
Whilst that of SiH4
= 35
The lesser the pKa the higher the Ka which means more acidic.
Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
Answer:
(☞ ಠ_ಠ)☞
Newton's. first law of motion states that "Everybody continues in its state of rest or uniform motion in a straight line unless it is acted by some external forces".
Explanation:
hope it helps
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tysm (≧▽≦)