Answer:
5.71 g
Explanation:
Step 1: Write the balanced equation
2 K + Cl₂ ⇒ 2 KCl
Step 2: Calculate the moles corresponding to 12.0 g of KCl
The molar mass of KCl is 74.55 g/mol.
12.0 g × 1 mol/74.55 g = 0.161 mol
Step 3: Calculate the moles of Cl₂ needed to produce 0.161 moles of KCl
The molar ratio of Cl₂ to KCl is 1:2. The moles of Cl₂ needed are 1/2 × 0.161 mol = 0.0805 mol
Step 4: Calculate the mass corresponding to 0.0805 moles of Cl₂
The molar mass of Cl₂ is 70.91 g/mol.
0.0805 mol × 70.91 g/mol = 5.71 g
Answer:
False
Explanation:
The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE
This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺
In order from the most likely to bind an oxygen to least likely;
3 bound o2, po2=100mmhg1 bound o2, po2=100mmhg3 bound o2, po2=40mmhg<span>1 bound o2, po2=40mmhg
</span>
Haemoglobin is more likely to bind oxygen if its other oxygen binding sites have already bound to an oxygen molecule. The higher the partial pressure of oxygen in the blood also makes it more likely that the hemoglobin will bind oxygen.
<span />
