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ser-zykov [4K]
3 years ago
15

You’re probably familiar with the sour taste of acidic lemon juice and the slippery feel of alkaline (basic) soap. In fact, thes

e characteristics were used to identify acids from bases long ago. Today, we understand much more about acidity and alkalinity, far beyond taste and feel. Acids are most simply defined as proton donors. They have a ph of below 7. They are substances that react with water to produce H3O+ (hydronium) ions (after all, H2O + one proton → H3O+ ). When you add an acid to water the water acts as a base, accepting protons from the acid. Bases are most simply defined as proton acceptors. They have a ph of above 7. They react with water to produce OH- (hydroxide) ions (H2O - one proton → OH-). When you add a base to water the water acts as an acid, donating protons to the base.1. Citric acid H3C6H5O7 is the substance that gives lemon juice and other citrus fruit juices a sour taste. How many grams of citric acid are present in 125mL of a 0.400M citric acid solution?
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
5 0

Answer:

9.606 grams of citric acid are present in 125 mL of a 0.400 M citric acid solution.

Explanation:

Molarity : It is defined as the number of moles of solute present in one liter of solution.  Mathematically written as:

Molarity=\frac{\text{Moles of solute}}{\text{volume of solution in L}}

Moles of citric acid  = n

Volume of the citric acid  solution = 125 mL  =125 × 0.001 L= 0.125 L

(1 mL = 0.001L)

Molarity of the citric acid  solution = 0.400 M

0.400 M=\frac{n}{0.12 5L}

n  = 0.400 M × 0.125 L = 0.05 moles

Mass of 0.05 moles of citric acid :

0.05 mol\times 192.12 g/mol=9.606 g

9.606 grams of citric acid are present in 125 mL of a 0.400 M citric acid solution.

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A powder contains FeSO4⋅7H2O (molar mass=278.01 g/mol), among other components. A 3.930 g sample of the powder was dissolved in
fredd [130]

Answer:

The mass of FeSO4*7H2O in the sample is 1.21 grams

Explanation:

<u>Step 1</u>: Calculate moles of Fe2O3

moles of Fe2O3 = mass of Fe2O3 / Molar mass of Fe2O3

moles of Fe2O3 = 0.348 grams / 159.69 g/mole = 0.00218 moles

<u>Step 2</u>: Calculate moles of Fe

4 Fe + 3O2 → 2Fe2O3

For 4 moles of Fe consumed there is 2 moles of Fe2O3 produced

This means it has a ratio 2:1

So 0.00218 moles of Fe2O3 produced , there is 2*0.00218 = 0.00436 moles of Fe consumed

<u>Step 3:</u> Calculate moles of FeSO4*7H2O

Fe + H2SO4 + 7H2O → FeSO4*7H20 + H2

For 1 mole of Fe consumed there is 1 mole of FeSO4*7H2O produced

This means for 0.00436 moles there is 0.00436 moles of Fe2SO4*H2O produced

<u>Step 4:</u> Calculate the mass of FeSO4*7H2O in the sample

mass of FeSO4*7H2O = 0.00436 moles * 278.01 g/mole = 1.212 g

The mass of FeSO4*7H2O in the sample is 1.21 grams

8 0
3 years ago
The enthalpy of vaporization of liquid water at 100°C is 2257 kJ/kg. Determine the enthalpy for apordato of iuod eeat capacity o
igomit [66]

Explanation:

The given data is as follows.

         T_{1} = 100^{o}C,       T_{2} = 10^{o}C

       \Delta H_{vap1} = 2257 kJ/kg,     \Delta H_{vap2} = ?

For water, C_{p} = 4.184 kJ/kg ^{o}C

Formula to calculate heat of vaporization is as follows.

  \Delta H_{vap1} - \Delta H_{vap2} = C_{p}(T_{1} - T_{2})

Hence, putting the values into the above formula as follows.

\Delta H_{vap1} - \Delta H_{vap2} = C_{p}(T_{1} - T_{2})

2257 kJ/kg - \Delta H_{vap2} = 4.184 kJ/kg ^{o}C (100 - 10)^{o}C

            \Delta H_{vap2} = 2257 kJ/kg - 376.56 kJ/kg

                                       = 1880.44 kJ/kg

Thus, we can conclude that enthalpy of liquid water at 10^{o}C is 1880.44 kJ/kg.

3 0
3 years ago
A 155g sample of copper was heated to 150.0 degrees Celsius, then placed into 250.0g water at 19.8 degrees Celsius. Calculate th
bogdanovich [222]
Once for the water and once for the copper. Set up a table that accounts for each of the variables you know, and then identify the ones you need to obtain. Give me a moment or two and I will work this out for you.

Okay, so like I said before, you will need to use the equation twice. Now, keep in mind that when the copper is placed in the water (the hot into the cold), there is a transfer of heat. This heat transfer is measured in Joules (J). So, the energy that the water gains is the same energy that the copper loses. This means that for your two equations, they can be set equal to each other, but the copper equation will have a negative sign in front to account for the energy it's losing to the water.

When set equal to each other, the equations should resemble something like this:
(cmΔt)H20 = -(cmΔt)Cu
(Cu is copper).

Remember, Δt is the final temperature minus the initial temperature (T2-T1). We are trying to find T2. Since we are submerging the copper into the water, we can assume that the final temperature at equilibrium is the same for both the copper and the water. At a thermodynamic equilibrium, there is no heat transfer because both materials are at the same temperature.

T2Cu = T2H20

Now, the algebra for this part of the problem is a bit confusing, so make sure you keep track of your variables. If done right, the algebra should work out so you have this:

T2 = ((cmT1)Cu + (cmT1)H20) / ((cm)H20 + (cm)Cu)
Insert the values for the variables. Once you plug and chug, your final answer should be
26.8 degrees Celsius.
3 0
3 years ago
Sodium sulfide reacts with hydrochloric acid to produce hydrosulfuric acid and sodium chloride. (you need to write and balance t
VikaD [51]

Answer:

2.5 moles of NaCl

Explanation:

The balanced chemical reaction equation is shown in the image. Since it takes 2 moles of Hydrochloric acid to form two moles of sodium. Chloride, then 2.5 moles of hydrochloric acid should also form 2.5 moles of sodium chloride according to the balanced reaction equation.

4 0
3 years ago
A sample of hydrated magnesium sulfate (MgSO4)
yaroslaw [1]

Answer:

MgSO4.7H2O

Explanation:

Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O

Mass of the hydrated salt (MgSO4.xH2O) = 12.845g

Mass of anhydrous salt (MgSO4) = 6.273g

Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g

Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:

Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x

Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt

18x/120 + 18x = 6.572/12.845

Cross multiply to express in linear form

18x x 12.845 = 6.572(120 + 18x)

231.21x = 788.64 + 118.296x

Collect like terms

231.21x — 118.296x = 788.64

112.914x = 788.64

Divide both side by 112.914

x = 788.64 /112.914

x = 7

Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O

6 0
3 years ago
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