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exis [7]
3 years ago
9

What is the third element in group nine?

Chemistry
1 answer:
Stels [109]3 years ago
3 0
The third element in group nine is: I<span>ridium</span>
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365,000,000 scientific notation
bekas [8.4K]
3.65 X 10 to the power of 8
6 0
3 years ago
g For the following reaction, 0.500 moles of silver nitrate are mixed with 0.285 moles of copper(II) chloride. What is the formu
scZoUnD [109]

Answer:

CuCl_2 is the formula for the limiting reagent.

Mass of silver chloride produced is 71.8 g.

Explanation:

CuCl_2+2AgNO_3\rightarrow 2AgCl+Cu(NO_3)_2

Moles of silver nitrate = 0.500 mol

Moles of copper(II) chloride = 0.285 mol

According to reaction, 2 moles of silver nitrate reacts with 1 mole of copper chloride , then 0.500 mole of silver nitrate will react with :

\frac{1}{2}\times 0.500 mol=0.250 mol of copper(II) chloride

As we can see that moles of copper(II) chloride will be reacting is 0.250 mol less than present moles of copper (II) chloride ,so this means that silver nitrate is limiting reagent.

And moles of silver chloride to be formed will depend upon silver nitrate.

According to reaction, 2 moles of silver nitrate gives 2 moles of silver chloride , then 0.500 mole of silver nitrate will give  :

\frac{2}{2}\times 0.500 mol=0.500 mol of silver chloride

Mass of silver chloride produced:

0.500 mol × 143.5 g/mol = 71.8 g

7 0
3 years ago
If 1.546 g of copper was used by a student at the start of the lab, and 0.732 g of copper were obtained at
mash [69]

Answer: Percent recovery is 47.34 %

Explanation:

Percent yield is defined as the ratio of experimental yiled to theoretical yield in terms of percentage.

{\text{ percent yield}}=\frac{\text{amount recovered}}{\text{total amount}}\times 100

Putting in the values we get:

{\text{ percent yield}}=\frac{0.732}{1.546}\times 100=47.34\%

Therefore, the percent recovery is 47.34 %

8 0
3 years ago
Which statement is true about molarity and percent by mass? (3 points)
lakkis [162]

Answer:  The correct option is, They are different units of concentration.

Explanation: Molarity : It is defined as the number of moles of solute present in one liter of solution.

5 0
3 years ago
Read 2 more answers
The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?
Vitek1552 [10]
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
     pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
     [OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
     NaA → Na+ + A- 
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
                             A- + H2O ⇌ HA + OH-
     Initial             0.0100            0       0
     Change        -x                    +x     +x
     Equilibrium    0.0100-x         x       x

We can now calculate the Kb for A-:
     Kb = [HA][OH-] / [A-] 
           = x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
     Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
     Ka = Kw / Kb 
           = 1.00x10^-14 / 1.00x10^-4
           = 1.00x10^-10
7 0
3 years ago
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