0.0760 m
do this by:
finding the moles of NaOH which will be <span>5.702 E -3 m
</span>
next find the moles of H3PO4 which will be <span>1.90 E -3 m</span><span>
calulcate </span>25 ml sample molarity = 0.07603 m, just put 0.0760<span>
</span>
Answer:
Explanation:
Iodine - 125
The atomic symbol of iodine is ¹²⁵₅₃ I
The symbol for iodine is I
The atomic number of iodine is 53,
and the atomic mass of iodine is 125 .
<u>The representation of the atomic symbol is as, the atomic mass is written in uppercase and the atomic number is written in lower case , followed by the symbol of the element .</u>
Iodine is a radio active element , used for many biological process .
It is the second largest -lived radioisotope of iodine .
The first is iodine-129 .
As in math or
?
/.........
Answer:
The correct answer is 0, 235 mol
Explanation:
We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol:
1 atm x 5, 25l = n x 0, 082 l atm / K mol x 273 K
n= 1 atm x 5, 25l /0, 082 l atm / K mol x 273 K
n= 0, 235 mol
<span>Let's assume
that the oxygen gas has ideal gas behavior.
Then we can use ideal gas formula,
PV = nRT</span>
Where, P is the pressure of the gas (Pa), V is the volume of the gas
(m³), n is the number of moles of gas (mol), R is the universal gas
constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.
<span>
P = 2.2 atm = 222915 Pa
V = 21 L = 21 x 10</span>⁻³ m³
n = ?
R = 8.314 J mol⁻¹ K⁻¹
<span>
T = 87 °C = 360 K
By substitution,
</span>222915 Pa x 21 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻<span>¹ x 360 K
n
= 1.56</span><span> mol</span>
<span>
Hence, 1.56 moles of the oxygen gas are </span><span>
left for you to breath.</span><span>
</span>
