<span> In this question (t½) of C-14 is 5730 years, which means that after 5730 years half of the sample would have decayed and half would be left as it is. After 5730 years ( first half life) 70 /2 = 35 mg decays and 35 g remains left. After another 5730 years ( two half lives or 11460 years) 35 /2 = 17.5mg decays and ...</span>Missing: <span>25g</span>
3.01 x 10^23 molecules (1 mole/6.203 x 10^23 molecules) = 0.500 moles<span>
</span>
Answer:
C. 3.74 g/L
Explanation:
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Also,
Moles = mass (m) / Molar mass (M)
Density (d) = Mass (m) / Volume (V)
So, the ideal gas equation can be written as:
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Molar mass of krypton gas = 83.8 g/mol
Applying the equation as:
1 atm × 83.8 g/mol = d × 0.0821 L.atm/K.mol × 273.15 K
⇒d = 3.74 g/L
<u>Answer:</u> The pH of the cleaning compound is 10.44
<u>Explanation:</u>
pH is defined as the negative logarithm of hydrogen or hydronium ion concentration that are present in a solution.
The equation representing pH of the solution follows:
We are given:
Putting values in above equation, we get:
Hence, the pH of the cleaning compound is 10.44
Answer:
1.01 amu
Explanation:
Published atomic mass = 1.00797amu => 1.01 amu (3 sig-figs)