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FrozenT [24]
1 year ago
6

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

esults? how?
Chemistry
2 answers:
Mnenie [13.5K]1 year ago
6 0
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
OlgaM077 [116]1 year ago
6 0

If the standard solutions had unknowingly been made up to be 0.0024 M AgNO₃ and 0.0040 M K₂CrO₄ the chemical equation would be:  

2AgNO₃ + K₂CrO₄ → Ag₂CrO₄ (s) + 2KNO₃

By mixing 0.0024 M AgNO₃ and 0.004 M K₂CrO₄ one would have Ag₂CrO₄ precipitated out.  

Thus, one is left with 0.0024 M KNO₃ mixed with (0.004 -0.0024 / 2) M = 0.0028 M of K₂CrO₄.  

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93.2 mL of a 2.03 M potassium fluoride (KF) solution
Marrrta [24]

Answer:

1.98 M

Explanation:

Given data

  • Initial volume (V₁): 93.2 mL
  • Initial concentration (C₁): 2.03 M
  • Volume of water added: 3.92 L

Step 1: Convert V₁ to liters

We will use the relationship 1 L = 1000 mL.

93.2mL \times \frac{1L}{1000mL} = 0.0932 L

Step 2: Calculate the final volume (V₂)

The final volume is the sum of the initial volume and the volume of water.

V_2 = 0.0932L + 3.92 L = 4.01L

Step 3: Calculate the final concentration (C₂)

We will use the dilution rule.

C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M

3 0
1 year ago
Read 2 more answers
A 92.8 gram sample of CuSO4•5H2O.(copper sulfate pentahydrate) is heated until water is released. How many grams of water were r
Ivan

Mass of water released =

92.8 g CuSO_{4}.5H_{2}O×\frac{5 * 18 g H_{2}O}{249.68 g CuSO_{4}.5H_{2}O}

= 33.45 g H_{2}O

7 0
1 year ago
Read 2 more answers
Consider four beakers labeled A, B, C, and D, each containing an aqueous solution and a solid piece of metal. Identity the beake
Anna71 [15]

Answer:

A Reaction

3. Pt(NO₃)₂(aq) + Cu(s)

4. Cr(s) + H₂SO₄(aq)

B Non Reaction

1. Mn(s) + Ca(NO₃)₂(aq)

2. KOH(aq) +  Fe(s)

Y > Q > W > Z > X

Explanation:

The first question is whether a reaction will occur base on the chemical equation below.

1. Mn(s) + Ca(NO₃)₂(aq)

2. KOH(aq) +  Fe(s)

3. Pt(NO₃)₂(aq) + Cu(s)

4. Cr(s) + H₂SO₄(aq)

Firstly, some element are more reactive than others , base on this criteria element can be arranged  base on it reactivity .

1. Mn(s) + Ca(NO₃)₂(aq)

This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.

2. KOH(aq) +  Fe(s)

The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.

3. Pt(NO₃)₂(aq) + Cu(s)

Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.

4. Cr(s) + H₂SO₄(aq)

Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.

Base on the reaction

Q + W+ Reaction occurs

Since the reaction occurred element Q is more reactive as it displace element w from it compound.  

X + Z+ No reaction

No reaction occurred because element x is less reactive than z therefore, it cannot displace z from it compound.

W + Z+ Reaction occurs

Element w is more reactive than z as it displaces z form it compound.

Q+ + Y Reaction occurs

Element Y is more reactive than element  Q as it displaces Q from it compound.

Therefore, the  order of reactivity from the most reactive to the least reactive will be Y > Q > W > Z > X

3 0
2 years ago
For each pair of gases, select one that most likely has the highest rate of effusion. Use the periodic table if necessary
brilliants [131]
You will do the highest one i think
4 0
2 years ago
Read 2 more answers
What is charge on the cation SnCl4
Rasek [7]
Its total charge is zero but for the elements:
Sn===> Sn4+ positive
Cl===> Cl- negative
7 0
1 year ago
Read 2 more answers
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