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fgiga [73]
3 years ago
8

300 moles of sodium nitrite are needed for a reaction. the solution is 0.450 m. how many ml are needed?

Chemistry
1 answer:
miv72 [106K]3 years ago
4 0

Answer:

\boxed{\text{667 000 mL}}

Explanation:

\text{Molar concentration} = \dfrac{\text{moles}}{\text{litres}}\\\\c = \dfrac{n}{v}

Data:

c = 0.450 mol·L⁻¹

n = 300 mol

Calculation:

\begin{array}{rcl}0.450 & = & \dfrac{300}{V}\\\\0.450V& = & 300\\\\V & = & \dfrac{300}{0.450}\\\\V & = & \text{667 L} =\textbf{667 000 mL}\end{array}\\\text{The volume of solution needed is }\boxed{\textbf{667 000 mL}}

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false

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Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
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Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

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3 years ago
Calculate the percent errorin a length measurementof 4.45cm if the correct value is 4.06
FinnZ [79.3K]

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9.6 %

Explanation:

<u>Step 1: How to define  percent error ? </u>

⇒ % error is the difference between a measured value and the known or accepted value

⇒Percent error is calculated using the following formula:

⇒%error  =  | Experimental value-theoretical value/theoretical value |  x100%

⇔ this can be written as well as : error = (| Experimental value/ theoretical value | - | Theoretical value / Theoretical value | ) x100%

<u>Step 2: Calculate % error</u>

In this case, this means :

%error = ( |(4.45 cm - 4.06cm ) / 4.06cm | ) x100%

%error = 0.096 x100%

%error =9.6 %

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3 years ago
Submit What is the solubility of Cd3(POA) 2 in water? (Ksp of Cd3(PO4)2 is 2.5 x 10-33) | 1 2 3 +/- . 0 x100
vesna_86 [32]

<u>Answer:</u> The solubility of Cd_3(PO_4)_2 in water is 1.18\times 10^{-7}mol/L

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Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}

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The expression for solubility constant for this reaction will be:

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Hence, the solubility of Cd_3(PO_4)_2 in water is 1.18\times 10^{-7}mol/L

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3 years ago
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