The pressure of the gas : 1.1685 atm
<h3>Further explanation</h3>
In general, the gas equation can be written
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
n=moles=1.5
V=volumes = 30 L
T=temperature=285 K
The pressure :
The solution would be like this for this specific problem:
Given:
pH of a 0.55 M hypobromous
acid (HBrO) at 25.0 °C = 4.48
[H+] = 10^-4.48 = 3.31 x
10^-5 M = [BrO-] <span>
Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>
To add, Hypobromous Acid does not require acid
adjustment, which is necessary for chlorine-based product and is stable and
effective in pH ranges of 5-9.<span>
</span>Hypobromous Acid combines with organic
compounds to form a bromamine. Chlorine also combines with the same organic
compounds to form a chloramine. <span>It is also
one of the least expensive intervention antimicrobial compounds available.</span>
Answer:
D.Salad
Explanation:
A mixture is a substance that combines two or more substance that are physically combined. A mixture can be easily seperated by a physical means . A mixture is the combination of various substances that are not chemically combined .Examples of mixtures are cake and salad . The cake contains various substances like egg, flour and sugar. This mixture can be easily separated into it individual entities by a physical means .
Base on the option the only mixture among the option is salad . Salad is a a mixtures that combines substances like tomatoes , fruits and vegetable. The individual substances that makes up the salad can be easily separated.
Sugar is a compound . Water is a compound as it contains two elements and potassium is an element.
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Answer:
The atomic mass of element is 65.5 amu.
Explanation:
Given data:
Abundance of X-63 = 50.000%
Atomic mass of X-63 = 63.00 amu
Atomic mass of X-68 = 68.00 amu
Atomic mass of element = ?
Solution:
Abundance of X-68 = 100-50 = 50%
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50×63)+(50×68) /100
Average atomic mass = 3150 + 3400 / 100
Average atomic mass = 6550 / 100
Average atomic mass = 65.5 amu.
The atomic mass of element is 65.5 amu.
