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Butoxors [25]
3 years ago
11

5 pos. 1. An excess of sodium sulfate was added to a 500. mL sample of polluted water. The

Chemistry
1 answer:
Ad libitum [116K]3 years ago
4 0

Answer:

m_{Pb^{2+}}=0.211gPb^{2+}

Explanation:

Hello!

In this case, according to the stoichiometry of the reaction, it is possible to evidence the 1:1 mole ratio between lead (II) ions and lead (II) sulfate precipitate; that is why we can compute the mass of lead (II) in the polluted water as shown below:

m_{Pb^{2+}}=308.88mgPbSO4*\frac{1gPbSO4}{1000mgPbSO4} *\frac{207.2gPb^{2+}}{303.26gPbSO4} \\\\m_{Pb^{2+}}=0.211gPb^{2+}

Best regards!

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3 years ago
Calculate the molarity of a solution obtained dissolving 10.0 g of cobalt(Ⅱ) bromide tetrahydrate in enough water to make 450 mL
Vladimir [108]

Answer:

<em><u>The molarity of the CoBr2•4H2O solution is  7.64 × 10-2 M</u></em>

Explanation:

Cobalt (II) bromide tetrahydrate

• Cobalt - A transition metal with Roman numeral (II) → charge: +2 → Co2+

• Bromide - anion from group 7A → -1 charge → symbol: Br-

• Tetrahydrate- tetra- means 4 and hydrate is H2O

The chemical formula of the compound is: CoBr2•4H2O

We then need to determine the number of moles of CoBr2•4H2O since this is the only information missing for us to find molarity. Notice that the volume of the solution is already given.

We’re given the mass of CoBr2•4H2O. We can use the molar mass of CoBr2•4H2O4 to find the moles.

•The molar mass of CoBr2•4H2O is:

CoBr2•4H2O  

1 Co x 58.93 g/mol Co = 58.93 g/mol

2 Br x 79.90 g/mol Br = 159.80 g/mol

8 H  x 1.008 g/mol H = 8.064 g/mol

4 O  x 16.00 g/mol O = 64.00 g/mol

________________________________________

                           Sum = <u>290.79 g/ mo</u>

The moles of CoBr2•4H2O is:

= 10.0 g CoBr2•4H2O x  \frac{ 1 mol  CoBr_2 . 4H_2O}{290.79 g CoBr_2 .  4H_2O}

= <u>0.0344  mol CoBr2•4H</u>

We know that the volume of the solution is 450 mL.

We can now calculate for molarity:

Convert mL to L → 1 mL = 10-3 L

Formula:

Molarity (M)= Mole of solute / Liters of solution

= 0.0344  mol CoBr2•4H  / 450 mL x 1 ml / 10^ -3 L

= 0.0764

=  7.64 × 10-2 mol/L

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f 37.4 grams of water decompose at 297 Kelvin and 1.30 atmospheres, how many liters of oxygen gas can be produced?
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2H2O=2H2+O2
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