Which of these will reflect light with the highest frequency?

If the standard solutions had unknowingly been made up to be 0.0024 m agno3 and 0.0040 m k2cro4, would this have affected your r

Mnenie [13.5K]

2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04

6 0

3 years ago

Read 2 more answers

Using what you have learned about magnetic forces, potential and kinetic energy. Describe the changes

weeeeeb [17]

Answer:

reversible reaction or covalent bonds

7 0

3 years ago

You are a chemist working in your laboratory. In your storage closet you have a cabinet that holds only samples of pure elements

Vladimir79 [104]

Answer:

Mercury

Explanation:

While it is true that most metals are solid at room temperature, mercury is liquid at room temperature hence mercury is often designated as the 'liquid metal'.

Thus, if i find a bottle on the shelf that has no solid in it, only liquid and i know that only pure metals are stored in that area of the laboratory, then i will quickly relabel it as mercury.

7 0

3 years ago

Which of the following is not an effect of acid rain? a. Corrosion of buildings c. Limestone corrosion b. Reduced cold tolerance

Novay_Z [31]

<span> d. Reduced visibility

Acid rain doesn't reduce visibility, smog does
</span>

3 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

3 years ago