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Mariulka [41]
3 years ago
14

Which of the following is represented in the highest percentage by volume in dry air?

Chemistry
1 answer:
Bogdan [553]3 years ago
8 0
Nitrogen is represented in the highest percentage by volume in dry air.
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After 11.5 days, 12.5% of a sample of radon-222 that originally weighed 42g remains. What is the half-life of this isotope?
Bond [772]
Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value.  The equation to describe the decay is
Nt=N0(1/2) ^{t/t(1/2)}  where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time.  So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days
5 0
3 years ago
Water can be formed in the following reaction: 2H2 + O2 —> 2H2O. If you have 12 moles of Hydrogen gas (H2), how many moles of
Kaylis [27]

Answer: 6

Explanation:

3 0
3 years ago
One beaker contains 19.93 mL, another contains 14.0 mL and a third contains 10.6 mL. What is the total volume of the three beake
Juli2301 [7.4K]

Answer:

\large \boxed{\text{44.5 mL}}

Explanation:

When adding or subtracting values, you must round your answer to the same "place" as the measurement with its last significant figure furthest to the left.  

That is, you round off to the same number of decimal places as the measurement with the fewest decimal places.

\begin{array}{r|r}19.9& \text{3 mL}\\14.0&\text{mL}\\10.6&\text{mL} \\\mathbf{44.5} &\textbf{3 mL}\\\end{array}

The measurements of 14.0 and 10.6 have one digit after the decimal point, so you round the sum to have only one digit to the right of the decimal.

The number to be dropped (3) is less than 5, so you drop it.

\text{The total volume of the three beakers is $\large \boxed{\textbf{44.5 mL}}$}

8 0
3 years ago
Why do neutral atoms form ions?
Marina86 [1]
Since they can still be unstable...nuetral atoms have the same amount of protons to electrons but to be stable they need to fill up there outer shell by gaining or losing electrons
8 0
3 years ago
What pressure, in atmospheres, is exerted on the body of a diver if he is 38 ft below the surface of the water when atmospheric
Talja [164]

The pressure of diver = atmospheric pressure + water pressure

atmospheric pressure = 750 mmHg (as given) = 750 / 760 atm = 0.987 atm

Water pressure is

P = hρg

where

h = height of water = 38 ft

1 ft = 0.3048

38 ft = 11.58 m

ρ = density  = 1000 Kg / m³

g = gravitational constant  = 9.81 m/s2

P = 11.58 X 1000 X 9.81 = 113599.8 Kg / m s^2 Or N /m^2

1 N / m^2 = 1 pa = 9.869 X 10^-6 atm

P = 113599.8 Pa = 1.12 atm

Total pressure = 1.12 + 0.987 atm = 2.107 atm = 2.1 atm (two significant figures)



5 0
3 years ago
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