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3 years ago

what is the concentration of 10.00ml of HBr if it takes 16.73 mL of a 0.253M LiOH solution to neutralize it?

1 answer:

4 0

The chemical reaction would be expressed as follows:

HBr + LiOH = LiBr + H2O

We are given the volumes and corresponding concentration to be used for the reaction. We use these values to solve for the concentration of the other reactant. We do as follows:

0.253 mol LiOH / L solution ( 0.01673 L ) ( 1 mol HBr / 1 mol LiOH ) = 0.00423 HBr needed

Concentration of HBr =0.00423mol / .010 L = <span>0.423 M HBr </span>

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Which list of elements consists of a metal, a metalloid, and a nonmetal?

Irina18 [472]

Answer:

Sn, Si, C

Explanation:

8 0

3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

2 years ago

Which of Graphs 1 correctly represents the relationship between the pressure and Kelvin temperature of a gas?

aliya0001 [1]

Answer:

Explanation:

Pressure is directly proportional to temperature

6 0

2 years ago

If your lawn is 21.0 ft wide and 20.0 ft long, and each square foot of lawn accumulates 1350 new snow flakes every minute. How m

klemol [59]

Let us start with the total area of the lawn. Area= width x length, ie, 21 x 20 = 420 sq. ft. Snow flakes per square foot per minute = 1350 So Snow flakes for 420 sq.feet per minute = 420 x 1350 = 567000. Snow flakes for 1 hour = 567000 x 60 = 34020000 (60 minutes) Weight of 34020000 snow flakes = 34020000 x 1.60 = 54432000mg. To convert it into kilograms, divide this number by 1000000 (1 kilogram = 1000000 milligrams) Thus 54432000/1000000 = 54.432 kilograms or 54 kilograms and 432 grams.

6 0

3 years ago

If the observed value for a density is 0.80 g/mL and the accepted value is 0.70 g/mL what is the percent error?

kvv77 [185]

Answer:

<h2>The answer is 14.29 %</h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual density = 0.70 g/mL

error = 0.8 - 0.7 = 0.1

So we have

$P(\%) = \frac{0.1}{0.7} \times 100 \\ = 14.285714...$

We have the final answer as

Hope this helps you

4 0

3 years ago