1. Calculate the percent oxygen in H2SO4.

A student places a 100.0°C piece of metal that weighs 85.5 g into 122 mL of 16.0°C water. If the final temperature is 20.2°C, wh

Musya8 [376]

Answer:

The specific heat of the metal is 0.314 J/g°C

Explanation:

Step 1: data given

Temperature of the piece of metal = 100.0 °C

Mass of the metal = 85.5 grams

Volume of water = 122 mL = 122 grams

Temperature of water = 16.0 °C

The final temperature of water = 20.2 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of metal

Heat gained= heat lost

Qgained = - Qlost

Qwater = -Qmetal

Q = m*c* ΔT

m(metal)*c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒m(metal) = mass of metal = 85.5 grams

⇒c(metal) = the specific heat of metal = TO BE DETERMINED

⇒ΔT(metal) = the change of temperature of metal = T2 - T1 = 20.2 - 100 °C = -79.8 °C

⇒m(water) = the mass of water = 122 grams

⇒c(water) = the specific heat of water = 4.184 J/g°C

⇒ΔT(water) = the change of temperature of metal = T2 - T1 = 20.2 - 16.0 °C = 4.2 °C

85.5 *c(metal) * -79.8 = -122 * 4.184 * 4.2

c(metal) * (-6822.9) = -2143.9

c(metal) = 0.314 J/g°C

The specific heat of the metal is 0.314 J/g°C

7 0

3 years ago

Potassium chlorate (kclo3) decomposes in a reaction described by this chemical equation:2kclo3(s) → 2kcl(s) + 3o2(g)if 36.0 gram

cricket20 [7]

<span>If 36 gm of potassium chlorate enter into the reaction, the total mass of the two products will still be 36 gm because if there is only one reactant, the mass of the compounds after the reaction will be same that reactant based on the law of conservation of matter.</span>

6 0

3 years ago

A 25.00 ml solution of sulfuric acid H2SO4 is titrated to phenolphthalein end point with 27.00 ml of 1.700 M KOH

Nastasia [14]

<h3>Answer:</h3>

0.918 M

<h3>Explanation:</h3>

Assuming the question requires we calculate the Molarity of sulfuric acid:

We are given:

Volume of the acid, H₂SO₄ = 25.00 ml
Volume of the base, KOH = 27.00 mL
Molarity of the base, KOH is 1.70 M

We can calculate the molarity of the acid using the following steps;

<h3>Step 1: Write the chemical equation for the reaction.</h3>

The reaction is an example of a neutralization reaction where a base reacts with an acid to form salt and water.

Therefore, the balanced equation will be;

H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)

<h3>Step 2: Determine the moles of the base, KOH </h3>

When given molarity and the volume of a solution, the number of moles can be calculated by multiplying molarity with volume.

Number of moles = Molarity × Volume

= 1.700 M × 0.027 L

= 0.0459 moles

Thus, moles of KOH used is 0.0459 moles

<h3>Step 3: Determine the number of moles of the Acid, H₂SO₄</h3>

From the reaction, 1 mole of the acid reacts with 2 moles of KOH

Therefore, the mole ratio of H₂SO₄ to KOH is 1 : 2

Thus, moles of H₂SO₄ = Moles of KOH ÷ 2

= 0.0459 moles ÷ 2

= 0.02295 moles

<h3>Step 4: Calculate the molarity of the Acid </h3>

Molarity is the concentration of a solution in moles per liter

Molarity = Moles ÷ Volume

Molarity of the acid = 0.02295 moles ÷ 0.025 L

= 0.918 M

Thus, the molarity of the acid, H₂SO₄ is 0.918 M

5 0

3 years ago

Given the equation C3H8(g) + O2(g) = CO2(g) + H2O(g) and that the enthalpies of formation for H2O(g) = -241.8 kJ/mol, CO2(g) = -

kari74 [83]

Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$ $\Delta H=-2220.1kJ/mol$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]$

$\Delta H_{C_3H_8}=72.4kJ/mol$

Therefore, the heat of formation of propane is 72.4 kJ/mol

7 0

3 years ago

25 pts<br> 3. Compare and contrast a dry cell with a voltaic cell by filling in the table.

Orlov [11]

Answer:

Half Cell (1)oxidation reaction (2) redox reaction

Anode----(1)and(2) electrochemical cell at which oxidation occurs

Cathode---(1)and(2)electrochemical cell at which reduction

Electrolytes----(1)ammonium chloride------

Oxidation Reaction --- (1)in Anode(2)in Anode

Reduction Reaction----(1)in Cathode(2)in cathode

Bridge---(1)Zinc–carbon battery(2) Galvanic cell

Relative size ------ (1)smaller--------(2)much larger

Explanation:

6 0

3 years ago