<em>K</em> = 2.4 × 10^(-72)
<em>Step 1</em>. Determine the <em>value of n
</em>
Zn^(2+) + 2e^(-) → Zn
2Cl^(-) → Cl_2 + 2e^(-)
Zn^(2+) + 2Cl^(-) → Zn + Cl_2
∴ <em>n</em> = 2
<em>Step 2</em>. Calculate <em>K</em>
log<em>K</em> = <em>nE</em>°/0.0592 V = [2 × (-2.12 V)]/0.0592 V = -71.62
<em>K</em> = 10^(-71.62) = 2.4 × 10^(-72)
<span>
<span> 250g H2SO4 x 1moleH2SO4/ 98g mole H2SO4 x 4moles Oxygen/1moleH2SO4 x 1moleOxgen/16g Oxygen
= 163.27g Oxygen
my figures aren't exact....so it's 160g </span>
</span>
For a carbonate containing a kind of cation, we have that Option 1 is the correct answer: ammonium.
<h3>What are Carbonates?</h3>
Generally, Any of two families of chemical substances generated from carbonic acid or carbon dioxide is referred to as carbonate.
In conclusion, Sodium, potassium, and ammonium are the three carbonates that are soluble.
Read more about Compound
brainly.com/question/704297
First, we have to get:
1- The heat required to increase T of ice from -50 to 0 °C:
according to q formula:
q1 = m*C*ΔT
when m is the mass of ice = mol * molar mass
= 1 mol * 18 mol/g
= 18 g
and C is the specific heat capacity of ice = 2.09 J/g-K
and ΔT change in temperature = 0- (-50) = 50°C
by substitution:
∴q1 = 18 g * 2.09 J/g-K *50°C
= 1881 J = 1.881 KJ
2- the heat required to melt this mass of ice is :
q2 = n*ΔHfus
when n is the number of moles of ice = 1 mol
and ΔHfus = 6.01 KJ/mol
by substitution:
q2 = 1 mol * 6.01 KJ/mol
= 6.01 KJ
3- the heat required to increase the water temperature from 0°C to 60 °C is:
q3 = m*C*ΔT
when m is the mass of water = 18 g
C is the specific heat capacity of water = 4.18 J/g-K
ΔT is the change of Temperature of water = 60°C - 0°C = 60°C
by substitution:
∴q3 = 18 g * 4.18 J/g-K * 60°C
= 4514 J = 4.514 KJ
∴the total change of enthalpy = q1+q2+q3
= 1.881 KJ +6.01 KJ + 4.514 KJ
= 12.405 KJ
Answer: 6.75 moles
Explanation:
This is a simple stoichiometry proboe. that I would set up like this:
(13.5 moles CuCI2) (1 mol I2 / 2 moles CuCi2)
That means you all you have to do for this problem is divide by 2 and cancel out the unit moles CuCI2, which leaves you with 6.75 moles I2.
Hope this helps :)
