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4 years ago

What is the formula of a compound in which element y form ccp lattice and atom x occupy 1/3rd of tetrahedral voids

Chemistry

1 answer:

Kisachek [45]4 years ago

5 0

Y : CCP : 4 atoms
X : tetrahedral voids would be 1/3 × 8 = 8/3

so formula would be Y12X8 or Y3X2 !!

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What is the pH for a 0.10 M HCI solution at 25 degreesCelsius?

11Alexandr11 [23.1K]

Answer:

1.0

Explanation:

Hydrochloric acid is a strong acid, that is, an acid that dissociates completely, according to the following reaction.

HCl(aq) → H⁺(aq) + Cl⁻(aq)

Then, the concentration of H⁺ will be equal to the initial concentration of the acid, i.e., 0.10 M.

We can calculate the pH using the following expression.

pH = -log [H⁺] = -log 0.10 = 1.0

3 0

3 years ago

If u answer this correctly I’ll mark you brainliest

olga2289 [7]

Answer:

6 is the right answer I know cause I like science

4 0

3 years ago

Read 2 more answers

a teacher uses a bow and arrow to demonstrate accuracy and precision. she shoots several arrows, aiming at the exact center of t

Vera_Pavlovna [14]

A teacher uses a bow and arrow to demonstrate accuracy and precision. she shoots several arrows, aiming at the exact center of the target each time. thedrawing below shows where her arrows hit the target.
 The statement that best describes her shots is
Her shots were neither accurate nor precise
hope it helps

7 0

3 years ago

A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h

Lostsunrise [7]

Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

5 0

3 years ago

What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i

Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314 (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0

3 years ago