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2 years ago

Pb(NO3)2+K2CrO4 --> PbCrO4+KNO3 How do you balance this?

Chemistry

1 answer:

belka [17]2 years ago

5 0

Answer:

Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃

Step-by-step explanation:

The unbalanced equation is

Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + KNO₃

Notice that the complex groups like NO₃ and CrO₄ stay the same on each side of the equation.

One way to simplify the balancing is to replace them with a single letter.

(a) For example, let X = NO₃ and Y =CrO₄. Then, the equation becomes

PbX₂ + K₂Y ⟶ PbY + KX

(b) You need 2X on the right, so put a 2 in front of KX.

PbX₂ + K₂Y ⟶ PbY + 2KX

Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃

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Copper metal (Cu) reacts with silver nitrate (AgNO3) in aqueous solution to form Ag and Cu(NO3)2. An excess of AgNO3 is present.

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Read 2 more answers

Calculate the mole fraction of the total ions in an aqueous solution prepared by dissolving 0.400 moles of CaI2 in 850.0 g of wa

Evgesh-ka [11]

Answer:

0.00840

Explanation:

The computation of the mole fraction is as follow:

As we know that

Molar mass = Number of grams ÷ number of moles

number of moles = Number of grams ÷ molar mass

Given that

Number of moles of CaI2 = 0.400

And, Molar mass of water = 18.0 g/mol

Now Number of moles of water is

= 850.0 g ÷ 18.0 g/mol

= 47.22 mol

And, Total number of moles is

= 0.400 + 47.22

= 47.62

So, Molar fraction of CaI2 is

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2 years ago

What is the predicted change in the boiling point of water when 1.50 g of

dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

⠀

$\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}$

⠀

The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.

⠀

$\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}$

⠀

Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

⠀

'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

⠀

'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

⠀

To find molality, we have to divide no. of moles of solute by weight of solution

⠀

While first we need to no. of moles

$\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072$

⠀

Now, we will find molality

⠀

$\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}$

⠀

$\textsf{ \large{ \underline{Now substituting the required values}}}$

⠀

$\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}$

⠀

Henceforth, the change in boiling point is 0.00735°C.

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1 year ago