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uranmaximum [27]
3 years ago
6

A solution containing a mixture of metal cations was treated with dilute hcl and a precipitate formed. the solution was filtered

and h2s was bubbled through the acidic solution. a precipitate again formed and was filtered off. then, the ph was raised to about 8 and h2s was again bubbled through the solution. this time, no precipitate formed. finally, the solution was treated with a sodium carbonate solution, which resulted in formation of a precipitate. which metal ions were definitely present, which were definitely absent, and which may or may not have been present in the original mixture?
Chemistry
1 answer:
likoan [24]3 years ago
6 0
If you really keep an eye on the flow chart, the only ions you can consider as being "Definitely not present" are: Cr3+, Fe3+, and Zn2+. The rest of the ions should be considered under "Possibly present", as we cannot conclude if any of the ions are "Definitely present". 
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Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74
stellarik [79]

Explanation:

Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

                                                                   = 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product (CH3COOH)^{2} gas formed are calculated as

0.010 mol CH3COOH * 1 mol (CH3COOH)^{2} / 2 mol CH3COOH

= 0.005 mol (CH3COOH)^{2}

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol (CH3COOH)^{2}

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure  :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

      = 0.74 atm * 0.015 mol / 0.020 mol

     = 0.555 atm

4 0
3 years ago
A sample of (NH4)2SO4 contains 0.750 mole. what is the mass of the sample​
steposvetlana [31]
The answer is 99.10.
5 0
2 years ago
How many capsules containing 75mg of Tamiflu could be produced from 155g of star anise.?
jeyben [28]
From the question you will find that:
one capsule of tamiflu is obtained from 2.6 g of star anise.
1 capsule   = 2.6 g tamiflu
? capsules = 155 g tamiflu
by cross multiplication = \frac{(1 x 155)}{2.6} = 59 capsules
7 0
3 years ago
Which of the following regions does NOT match its description?
Fantom [35]

The region that does not match its description is the fourth answer choice which is Abdominal region: spine

  • For the first answer choice - Cranial region: head

The cranial region encompasses the upper part of the head.

∴ The first answer choice matches its description.

  • For the second answer choice - Axillary region: armpits

The axillary region is an anatomical region under the shoulder joint where the arm connects to the shoulder. Therefore, it encompasses the armpits.

∴ The second answer choice matches its description.

  • For the third answer choice - Thoracic region: chest

The thoracic region runs from the base of the neck down to the abdomen. Therefore, it encompasses the chest

∴ The third answer choice matches its description

  • For the fourth answer choice - Abdominal region: spine

The abdominal region is divided into four quadrants which include are

1. Right upper quadrant fossa (RUQ)

2. Right lower quadrant fossa (RLQ)

3. Left lower quadrant fossa (LLQ)

4. Left upper quadrant fossa (LUQ)

It is also divided into nine (9) areas, all of which does not include the spine.

∴ The fourth answer choice does NOT match its description.

Hence, the region that does not match its description is the fourth answer choice which is Abdominal region: spine

Learn more here: brainly.com/question/11504303

8 0
1 year ago
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