Answer:
9g/cm^3 is the density
Explanation:
P = m/V
P = 18/2 = 9g/cm^3
(This is more of a physics question than chem btw)
Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g
Answer:
0.404M
Explanation:
...<em>To make exactly 100.0mL of solution...</em>
Molar concentration is defined as the amount of moles of a solute (In this case, nitrate ion, NO₃⁻) in 1 L of solution.
To solve this question we need to convert the mass of Fe(NO₃)₃ to moles. As 1 mole of Fe(NO₃)₃ contains 3 moles of nitrate ion we can find moles of nitrate ion in 100.0mL of solution, and we can solve the amount of moles per liter:
<em>Moles Fe(NO₃)₃ -Molar mass: 241.86g/mol-:</em>
3.26g * (1mol / 241.86g) =
0.01348 moles Fe(NO₃)₃ * (3 moles of NO₃⁻ / 1mole Fe(NO₃)₃) =
<em>0.0404 moles of NO₃⁻</em>
In 100mL = 0.1L, the molar concentration is:
0.0404 moles of NO₃⁻ / 0.100L =
<h3>0.404M</h3>
2.2311 moles of gas are there in a 50. 0 l container at 22. 0 °c and 825 torrs.
<h3>What is an ideal gas?</h3>
An Ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Assuming the gas is ideal, we can solve this problem by using the following equation:
PV = nRT
Where:
P = 825 torr ⇒ 825 / 760 = 1.08 atm
V = 50 L
n = ?
R = 0.082 atm·L·mol⁻¹·K⁻¹
T = 22 °C ⇒ 22 + 273.16 = 295.16 K
We input the data:
1.08 atm x 50 L = n x 0.082 atm·L·mol⁻¹·K⁻¹ x 295.16 K
And solve for n:
24.20312
n = 2.2311 mol
Hence, 2.2311 moles of gas are there in a 50. 0 L container at 22. 0 °c and 825 torrs.
Learn more about ideal gas here:
brainly.com/question/23580857
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