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3 years ago

Please help will give brainliest

Chemistry

1 answer:

snow_lady [41]3 years ago

5 0

Answer:

1st one is...

Explanation:

The 1st one is change in size or shape. 2nd one is formation of precipitate. 3rd one is physical change. 4th one is formation of gas.

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A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is

Airida [17]

Answer:

the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C

Explanation:

Given the data in the question;

First we calculate the Volume of the steel cylinder;

V = πr²h

radius r = Diameter / 2 = 27 cm / 2 = 13.5 cm

height h = 32.4 cm

so we substitute

V = π × ( 13.5 cm )² × 32.4 cm

V = π × 182.25 cm × 32.4 cm

V = 18550.79 cm³

V = 18.551 L

given that; maximum safe pressure P = 3.10 MPa = 30.5946 atm

vessel contains 0.218kg or 218 gram of carbon monoxide gas

molar mass of carbon monoxide gas is 28.010 g/mol

moles of carbon monoxide gas n = 218 gram / 28.010 g/mol = 7.7829 mol

we know that;

PV = nRT

solve for T

T = PV / nR

we know that gas constant R = 0.0820574 L•atm•mol⁻¹ K⁻¹

so we substitute

T = ( 30.5946 × 18.551 ) / ( 7.7829 × 0.082 )

T = 567.5604 / 0.6381978

T = 889.317387 K

T = ( 889.317387 - 273.15 ) °C

T = 616.167 ≈ 616 °C { 3 significant digits }

Therefore, the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C

6 0

3 years ago

The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper

LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

$t_{1/2}=\frac{1}{k\times a_0}$

k = rate constant =?

$a_0$ = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

a) Initial concentration of XY = $a_o=0.100 M$

Rate constant of the reaction = k = $6.96\times 10^{-3} M^{-1} s^{-1}$

Half life of the reaction is:

$t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}$

$=1,436.78 s$

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

$\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}$

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

$\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = $6.20\times 10^{-2} M$

$\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become $6.20\times 10^{-2} M$ .

e) Initial concentration of $SO_2Cl_2= 0.050 M$

Final concentration after time t = x

t = 55.0 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0

3 years ago

Write the electron configuration

ycow [4]

Answer:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

8 0

3 years ago

Select the properties of the Sn2 reaction mechanism. a. stereospecific: 100% inversion of configuration at the reaction site b.

ivann1987 [24]

a. stereospecific: 100% inversion of configuration at the reaction site

3 0

3 years ago

1 pts

soldier1979 [14.2K]

Answer:

144 g

Explanation:

Use the mole ratio of 4 mol CO2 for every 9 mol O2 to convert from mol O2 to mol CO2. Then use the molar mass of CO2 to convert from mol of CO2 to grams of CO2.

7.34 mol O2 • (4 mol CO2 / 9 mol O2) • (44.01 g CO2 / 1 mol CO2) = 144 g CO2

5 0

3 years ago

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