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2 years ago

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A 9.86 g sample of an aqueous solution of hydroiodic acid contains an unknown amount of the acid. If 27.5 mL of 0.337 M potassiu

m hydroxide are required to neutralize the hydroiodic acid, what is the percent by mass of hydroiodic acid in the mixture?
% by mass

1 answer:

Aneli [31]2 years ago

7 0

Answer:

88.3% of hydroiodic acid is found

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Helppppp asaapppppp plzzzzzz

Gala2k [10]

Answer:

Alright the very first thing you need to do is balance the equation:

2HCl + Na2CO3 -----> 2NaCl + CO2 + H2O

Now we need to find the limiting reactant by converting the volume to moles of both HCl and Na2CO3.

Volume x Concentration/molarity = moles

0.235L x 0.6 M = 0.141 moles / molar ratio of 2 = 0.0705 moles of HCl

0.094L x 0.75 M = 0.0705 moles /molar ratio of 1 = 0.0705 moles of Na2CO3

Since both of the moles are equal, it means the entire reaction is complete (while the identification of limiting reactant may seem like an unnecessary step, it's quite essential in stoichiometry, so keep an eye out) and there is no excess of any reactant.

Now we know that the product we want to calculate is aqueous so, following the law of conservation of mass, we should add both volumes together to calculate how much volume we could get for NaCl.

0.235 + 0.094 = 0.329L of NaCl

Now we apply the C1V1 = C2V2 equation using the concentration and volume of Na2CO3 because it's molar ratio is one to one to NaCl (You can also use HCL, but you have to divide their moles by 2 for the molar ratio) and the volume we just calculated for NaCl.

(0.75M) x (0.094L) = C2 x (0.329L)

Rearrange equation to solve for C2:

<u>(0.75M) x (0.094L)</u> = C2

(0.329L)

C2 = 0.214 M (Rounded)

<u>When the reaction is finished, the NaCl solution will have a molarity concentration of 0.214 M.</u>

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7 0

2 years ago

I have these answered, but I'd like someone to check them for me? Plz and thanks :)

zlopas [31]

1. The third option is the least soluble in water because it is the chain with the most number of hydrocarbons. Next is the second option while the first one is the most soluble.
2. Statements 1 and 2 are true. The third option is not true all the time because it depends on the structure of the compound.

4 0

3 years ago

In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in

RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 $m^{3}$

Density of water = 1000 $kg/m^{3}$

Therefore, mass of water = Density × Volume

= $1000 kg/m^{3} \times 0.25 m^{3}$

= 250 kg

Initial Temperature of water ( $T_{1}$ ) = $20^{o}C$

Final temperature of water = $140^{o}C$

Heat of vaporization of water ( $dH_{v}$ ) at $140^{o}C$ is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point ( $140^{o}C$ ) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

= 62.5 (kg) × 2133 (kJ/kg)

= $133.3 \times 10^{3}$ kJ

All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = $\frac{133.3 \times 10^{3}kJ}{3600 s}$ = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

= 125520 kJ

Time required = Heat required / Power rating

= $\frac{125520}{37}$

= 3392 sec

Time required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 0.25 $m^{3}$ water is calculated as follows.

$\frac{3392 sec}{60 sec/min}$

= 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0

3 years ago

What is the unit of measurement for mass?

Licemer1 [7]

Grams. It is a smaller unit.

4 0

3 years ago

Using the molarity equation, calculate how many grams of salt you need in order to make 80ml of a 2M solution (MW = 58.44g/mole)

ollegr [7]

Answer:

9.35g

Explanation:

The molarity equation establishes that:

$\textrm{molarity}=\frac{\textrm{moles of solute}}{\textrm{liters of solution}}$

So, we have information about molarity (2M) and volume (80 ml=0.08 l), with that, we can find the moles of solute:

$\textrm{moles of solute}=\textrm{molarity}*\textrm{liters of solution}$

$\textrm{moles of solute}= 0.08 \textrm{ l} *2\textrm{ M} = 0.16 \textrm{ mol}$

The mathematical equation that establishes the relationship between molar weight, mass and moles is:

$\textrm{molar weight}= \frac{\textrm{mass}}{\textrm{moles}}$

$\textrm{MW}= \frac{\textrm{m}}{\textrm{n}}$

We have MW (58.44g/mole) and n (0.16 mol), and we need to find m (grams of salt needed) to solve the problem:

$\textrm{m} = \textrm{MW * n}= 58.44\frac{\textrm{g}}{\textrm{mol}} * 0.16 \textrm{ mol} = 9.35 \textrm{ g}$

8 0

3 years ago