Answer:
14.2L at STP
Explanation:
Based on the problem, 2 moles of NH3 produce 6 moles of HF. To solve this question we have to convert the mass of NH3 to moles. With the chemical equation find the moles of HF and using PV = nRT find the liters of HF:
<em>Moles NH3 -Molar mass: 17.031g/mol-</em>
3.6g NH3 * (1mol / 17.031g) = 0.211 moles NH3
<em>Moles HF:</em>
0.211 moles NH3 * (6mol HF / 2mol NH3) = 0.634 moles HF
<em>Volume HF</em>
PV = nRT; V = nRT/P
<em>Where V is volume in liters, n are moles of the gas = 0.634 moles, R is gas constant = 0.082atmL/molK, T is absolute temperature = 273.15K at STP and P is pressure = 1atm at STP.</em>
Replacing:
V = 0.634moles*0.082atmL/molK*273.15K / 1atm
V = 14.2L at STP
Answer:
1. [OH⁻] = 0.30 M ; 2. [OH⁻] = 1.54x10⁻⁶M ; 3. [OH⁻] = 1.32x10⁻¹³M
Explanation:
Remember the rule:
pH + pOH = 14
pOH = 14 - pH
10*⁻pOH (you have to elevate 10, to -pOH)
10*⁻pOH = [OH⁻]
1. 14 - 13.48 = 0.52
10⁻⁰°⁵² = 0.30
2. 14 - 8.19 = 5.81
10⁻⁵°⁸¹ = 1.54x10⁻⁶
3. 14 - 2.12 = 12.88
10⁻¹²°⁸⁸ = 1.32x10⁻¹³
They must have the same number of protons
Answer:
Value = 1.80 g/cm³ (Approx)
Explanation:
Given:
Computation:
Value = 1.80 g/cm³ (Approx)
Answer:
option D is correct
D. This solution is a good buffer.
Explanation:
TRIS (HOCH
)
CNH
if TRIS is react with HCL it will form salt
(HOCH)CNH + HCL ⇆ (HOCH)NHCL
Let the reference volume is 100
Mole of TRIS is = 100 × 0.2 = 20
Mole of HCL is = 100 × 0.1 = 10
In the reaction all of the HCL will Consumed,10 moles of the salt will form
and 10 mole of TRIS will left
hence , Final product will be salt +TRIS(9 base)
H = Pk
+ log (base/ acid)
8.3 + log(10/10)
8.3
