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lesya [120]
3 years ago
5

How many liters of oxygen gas, at standard temperature and pressure, will react with 35.4 grams of calcium metal? Show all of th

e work used to solve this problem. 2Ca + O2 yields 2CaO
Chemistry
1 answer:
maks197457 [2]3 years ago
3 0
9.9 liters is your answer

Hope this helps ;)
You might be interested in
Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result
Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0
2 years ago
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
KIM [24]

Answer:

ΔH = -470.4kJ

Explanation:

It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:

1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ

2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ

6 times the reaction 1.

6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ

This reaction + 2:

6CaC2(s) + 3CO2(g) + 16H2O(l) →  + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ

As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:

6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) →  + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5

<h3>ΔH = -470.4kJ</h3>

4 0
2 years ago
A flashlight battery is an example of a
kompoz [17]
<span>Dry cell battery

When an automotive battery is fully charged, the sulfuric acid and water mixture will have a specific gravity of about 1.3. Specific gravity is actually the difference in the weight of water in comparison to a specific fluid. It is measured by a hydrometer. The amount of charge in the battery is normally measured by the specific gravity of the battery. The specific gravity of water is 1 and anything less than one is considered less dense while anything that has a specific gravity of more than 1 is considered more dense than water. </span>
7 0
3 years ago
In a solution with a pH of 3.0, the color of
Alinara [238K]

Answer:

A) litmus is red

Explanation:

To answer this question, it can be helpful to have the color charts. Litmus, phenolphthalein and methyl orange are ways to test the pH of a substance.

<u>Litmus paper</u>

Litmus can tell you if a substance is an acid or a base. You need to put the substance on both red litmus and blue litmus paper.

pH < 7: both papers are red. 3.0 is less than 7.

pH = 7: none of them change color

pH > 7: both papers are blue

<u>Phenolphthalein</u>

When this indicator is added to a substance, the result is either colorless or pink.

0 < pH ≤ 7: colorless. The color is not red or blue for pH 3.0.

pH > 7: pink

<u>Methyl orange</u>

0 < pH < 4: red. The color is not yellow if the pH is 3.0.

4 ≤ pH < 5: orange

pH ≥ 5: yellow

5 0
2 years ago
What is the molar solubility of agcl in 0. 30 m nacl at 25°C. ksp for agcl is 1. 77 × 10^-10.
Rzqust [24]

Molar solubility of AgCl will be  0.59 ×  10^{-10} M.

The amount of a chemical that can dissolve in one liter of a solution before reaching saturation is known as its molar solubility. This implies that the quantity of a substance it can disintegrate in a solution even before the solution becomes saturated with that particular substance is determined by its molar solubility.

A compound's molar solubility would be the measure of how many moles of such a compound must dissolve to produce one liter of saturated solution. The molar solubility unit will be mol L-1.

Calculation of molar solubility:

Given data:

M = 0.30 M

K_{sp} = 1.77 × 10^{-10}

The reaction can be written as:

AgCl ⇔ Ag^{+} + Cl^{-}

s            s         (s+0.30)

K_{sp}  = [Ag^{+} ]+ [Cl^{-}]

1.77 × 10^{-10} = s (0.30)

s = 1.77 × 10^{-10}  / 0.3

s = 0.59 ×  10^{-10} M

Therefore, molar solubility of AgCl will be  0.59 ×  10^{-10} M.

To know more about molar solubility

brainly.com/question/16243859

#SPJ4

3 0
1 year ago
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