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Ksivusya [100]
3 years ago
9

Calculating the volume of 0.05mol/dm3 KOH is required to neutralise 25.0cm3 of 0.0150mol/dm3 HNO3

Chemistry
1 answer:
kvasek [131]3 years ago
8 0

Answer:

You first need to construct a balanced chemical equation to describe the reaction:

KOH + HNO3 ---------> KNO3 + H2O

Work out the no. moles of HNO3 being neutralized:

Moles = Volume x Concentration = (25/1000) x 0.0150 = 0.000375 moles

From the balanced equation the molar ratio of KOH to HNO3 is 1:1 so you also need 0.000375 moles of KOH to neutralise the nitric acid

Now you can work out the volume of KOH required:

Volume = Moles/Concentration = (0.000375)/0.05 = 0.0075 dm^3 = 7.5 cm^3

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What's the molar mass of Li3p
spayn [35]

Answer:

51.79g Li₃P.

Explanation:

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4 0
3 years ago
How many grams of Cl are in 525g of CaCl2
Tema [17]

First we determine the moles CaCl2 present:

525g / (110.9g/mole) = 4.73 moles CaCl2 present 

Based on stoichiometry, there are 2 moles of Cl for every mole of CaCl2:<span>
(2moles Cl / 1mole CaCl2) x 4.73 moles CaCl2 = 9.47 moles Cl </span>

Get the mass:<span>
<span>9.47moles Cl x 35.45g/mole = 335.64 g Cl</span></span>

8 0
3 years ago
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fomenos

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5 0
2 years ago
. How many milliliters of 0.20 M HCl are needed to exactly neutralize 40. milliliters of 0.40 M KOH
Anuta_ua [19.1K]

Answer:

V_{HCl}=80mL

Explanation:

Hello,

In this case, for the given reactants we identify the following chemical reaction:

KOH+HCl\rightarrow KCl+H_2O

Thus, we evidence a 1:1 molar ratio between KOH and HCl, therefore, for the complete neutralization we have equal number of moles, that in terms of molarities and volumes become:

n_{HCl}=n_{KOH}\\\\M_{HCl}V_{HCl}=M_{KOH}V_{KOH}

Hence, we compute the volume of HCl as shown below:

V_{HCl}=\frac{M_{KOH}V_{KOH}}{M_{HCl}} =\frac{0.40M*40mL}{0.20M} \\\\V_{HCl}=80mL

Best regards.

5 0
3 years ago
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