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3 years ago

4. Rock is broken down as a result of chemical reactions;

Chemistry

2 answers:

tekilochka [14]3 years ago

7 0

Answer:

Explanation:

CHEMICALS WEATHER IN CHEMICAL WEATHERING

Alexandra [31]3 years ago

4 0

Answer:

Explanation:

it's called chemical weathering

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What is the body's first line of defense against bacteria and viruses? saliva, mucus, and hairs antibodies lymphocytes lymph nod

abruzzese [7]

Well your skin is the first line of defense

5 0

3 years ago

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4). One mole of monoclinic sulfur at 25C was placed in a constant-pressure calorimeter whose heat capacity (C) was 1620 J/K. T

andre [41]

Answer: The enthalpy change of the reaction is -243 J/mol

Explanation:

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

$q=c\times \Delta T$

where,

c = heat capacity of calorimeter = 1620 J/K

$\Delta T$ = change in temperature = $0.150^oC=0.150K$ (Change remains same)

Putting values in above equation, we get:

$q=1620J/K\times 0.15K=243J$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

$S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}$

We are given:

Moles of monoclinic sulfur = 1 mole

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -243 J

n = number of moles = 1 mole

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol$

Hence, the enthalpy change of the reaction is -243 J/mol

8 0

3 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

which of the following is a product formed whrb NO2 and H2O react together A O2 B HNO3 C H2 D P(OH)2

marshall27 [118]

The answer to your question is HNO3

8 0

3 years ago

Read 2 more answers

An iron block of mass 18 kg is heated from 285 K to 318 K. If 267.3 kJ is required, what is the specific heat of iron? A. 450.00

valkas [14]

Answer:

Option A. 450.00

Explanation:

1) Data:

a) m = 18 kg

b) T₁ = 285 K

c) T₂ = 318 K

d) Q = 267.3 kJ

e) S = ?

2) Principles and equations

The specific heat of a substance is the amount of heat energy absorbed to increase the temperature of certain amount (gram, kg, or moles, depending on the definition or units) of the substance in 1 ° C or 1 K.

The mathematical relation between the specific heat and the heat energy absorbed is:

Q = m × S × ΔT

Where,

Q is the heat absorbed,
S is the specific heat, and
ΔT is the temperature increase (T₂ - T₁)

3) Solution:

a) Substitute the data into the equation:

267.3 kJ = 18 kg × S × (318 K - 285 K)

b) Solve for S and compute:

S = 267.3 kJ / (18 kg × 33 K) = 0.45 kJ / (Kg . K)

The options have not units, but I notice that the first answer is 1,000 times the answer I obtained, so I will make a conversion of units.

c) Convert to J /( kg . k):

0.45 kJ / (Kg . K) × 1,000 J / kJ = 450 J / (kg . K)

Now we can see that the option A is is the answer, assuming the units.

6 0

3 years ago