Explanation:
so moles = mass ÷ mr (1+ 79.9)
so 10.00g ÷ 80.9
which is 0.1236093943
so to 3 S.F is 0.124 moles
also there is 1 to 1 ratio for LiOH to HBr
hope this helps :)
Answer:
C) Q < K, reaction will make more products
Explanation:
- 1/8 S8(s) + 3 F2(g) ↔ SF6(g)
∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³
∴ Q = [ SF6 ] / [ F2 ]³
∴ [ SF6 ] = 2 mol/L
∴ [ F2 ] = 2 mol/L
⇒ Q = ( 2 ) / ( 2³)
⇒ Q = 0.25
⇒ Q < K, reaction will make more products
Another advantage of advantage of using a microspectrophotometer to analyze fibers asides not causing damage to the sample is that the sample can be quite small.
<h3>What is a microspectrophotometer?</h3>
Microspectrophotometry is a biological technique used to measure the absorption or transmission spectrum of a solid or liquid material in either transmitted or reflected light.
Microspectrophotometry can also measure the emission of light by a sample, which is usually small as the micro implies.
One advantage of microspectrophotometry is that the sample does not get damaged. However,
However, another advantage of advantage of using a microspectrophotometer to analyze fibers asides not causing damage to the sample is that the sample can be quite small.
Learn more about microspectrophotometry at: brainly.com/question/5832827
The formula for the compounds in the reaction are as follows with the respective states
Carbon monoxide - CO (g)
hydrogen - H₂ (g)
methane - CH₄(g)
water - H₂O (l)
reaction of carbon monoxide with hydrogen gas gives rise to methane and water
the balanced chemical equation for the above reaction is as follows
CO(g) + 3H₂(g) --> CH₄(g) + H₂O(l)
Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol