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3 years ago

This diagram shows a marble with a mass of 3.8 grams (g) that was placed into 10 milliliters (mL) of water.

Chemistry

1 answer:

ryzh [129]3 years ago

3 0

Complete Question:

This diagram shows a marble with a mass of 3.8 grams (g) that was placed into 10 milliliters (mL) of water. Using the formula V M D = , what is the density of the marble?

(See attachment for full diagram)

Answer:

1.27 g/cm³

Explanation:

First, find the volume of rock:

Volume of rock = volume of water after rock was placed - volume of water before rock was placed

Volume of rock = 13 - 10 = 3ml

Density of rock = grams of rock per 1 cm³

Note: 1 ml = 1 cm³

Let x represent amount of rock per 1 cm³

Thus,

3.8g = 3 cm³

x = 1 cm³

Cross multiply

1*3.8 = 3*x

3.8 = 3x

3.8/3 = 3x/3

1.27 = x

Density of rock = 1.27 g/cm³

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Rama09 [41]

Explanation:

C: $CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$ ΔH=+178 kJ/mol

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

$A+\text{heat}\rightleftharpoons B$

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Treat heat as a reactant and on increasing a reactant at equilibrium, shifts the reaction in the forward direction.

Increase temperature → increase in heat → forward direction

Decrease temperature → decease in heat → backward direction

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System C - Decrease temperature : Reaction will move backward

D: $PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$ ΔH=−88 kJ/mol

The total enthalpy of the reaction comes out to be negative .

The temperature of the surrounding will increase.

For an exothermic reaction, heat is released during a chemical reaction and is written on the product side.

$A\rightleftharpoons B+\text{ heat}$

Treat heat as a product and on increasing a product at equilibrium, shifts the reaction in the backward direction.

Increase temperature → increase in heat → backward direction

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Which atom would be expected to have a half-filled 3d subshell?

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If the two atoms participating in a bond have an electron negativity difference of 1.9 , what type of of bond would they be part

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When bonded atoms have greater electronegativity difference i.e 2 or greater than two the bond is ionic because electron is transfer from low electronegative atom to highest electronegative atom.

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In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.

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3 years ago

Read 2 more answers

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

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