Ethylene glycol (antifreeze) has a density of 1.11 g/cm3.

if 50.0 ml of a 1.00 M HCl solution and 45.0 ml of a 1.00 M NaOH solution are combined, which reagent is limiting?

Setler79 [48]

The answer to this question would be: NaOH solution

Limiting reagent is the reagent that totally consumed in the reaction. To answer this question you need to know the equation for the chemical reaction. The equation would be:
HCl + NAOH = H2O + NACl

All of the coefficient in the reaction is 1, which mean 1 HCl will react to 1 NaOH. In this question, the concentration of HCl and NaOH is same so the amount of the molecule can be reflected by the volume. NaOH has lower volume compared to HCl so it is clear that NaOH will be totally consumed in this reaction.

3 0

3 years ago

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Convert the pressure 0.8 atm to kPa.

Ludmilka [50]

Answer:

81.0

Explanation:

mutiply the value by 101.325 to get the answer a. 81.1

3 0

3 years ago

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How much water must be added to make a 1.0 M solution from 250. mL of a 2.75 M solution?

Stels [109]

700 mL ....................

4 0

3 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$