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3 years ago

What pressure (in atm) will 0.44 moles of co2 exert in a 2.6 l container at 25Â°c?

1 answer:

6 0

We can use the ideal gas law equation to find the pressure
PV = nRTwhere
P - pressure
V - volume - 2.6 x 10⁻³ m³
n - number of moles - 0.44 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values into the equation,
P x 2.6 x 10⁻³ m³ = 0.44 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 419 281.41 Pa
101 325 Pa is equivalent to 1 atm
Therefore 419 281.41 Pa - 1/ 101 325 x 419 281.41 = 4.13 atm
Pressure is 4.13 atm

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9 x 1025 m<br> How many significant figures are in<br> this measurement?

sattari [20]

Answer: 9000 - 1 significant figure

Explanation: Since you are multiplying, the number with the least amount of significant figures determines the number of significant figures in the answer. The number 9 has 1 significant figure and 1025 and has 4 significant figures so 9 has the fewest significant figures, meaning your answer will be 1 significant figure.

8 0

3 years ago

Ethylenediamine (en) forms an octahedral complex with Ni2+(aq) with the formula [Ni(en)3]2+. Ni2+(aq) + 3 en ⇌ [Ni(en)3]2+(aq) K

yawa3891 [41]

Answer:

$3.125\times 10^{-19} mol/L$ is the concentration of $Ni^{2+}(aq)$ in the solution.

Explanation:

$Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)$

Concentration of nickel ion = $[Ni^{2+}]=x$

Concentration of nickel complex= $[[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L$

Concentration of ethylenediamine = $[en]=\frac{0.80 mol}{2 L}=0.40 mol/L$

The formation constant of the complex = $K_f=4.0\times 10^{18}$

The expression of formation constant is given as:

$K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}$

$4.0\times 10^{18}=\frac{0.08 mol/L}{x\times (0.40 mol/L)^3}$

$x=\frac{0.08 mol/L}{4.0\times 10^{18}\times (0.40 mol/L)^3}$

$x=3.125\times 10^{-19} mol/L$

$3.125\times 10^{-19} mol/L$ is the concentration of $Ni^{2+}(aq)$ in the solution.

6 0

4 years ago

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Serga [27]

Answer:

the answer is B

Explanation:

5 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Solving the Rydberg equation for energy change gives ΔE = R[infinity]hc [ 1 n12 − 1 n22 ] where the

icang [17]

Answer:

Explanation:

Utilizing Rydber's equation:

ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have

n=1 to n= infinity

ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)

n= 3 to n= infinity

ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J

b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .

1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹

1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m

λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm

7 0

3 years ago

Read 2 more answers

. Monthly measurements of atmospheric CO2 concentration at Mauna Loa began in March 1958. The average CO2 concentration for that

Harrizon [31]

Answer:

$Increase=98.8ppm$

$Average\ increase/year=1.594\frac{ppm}{year}$

Explanation:

Hello,

In this case, since the nowadays concentration of CO2 is 414.51 ppm and the concentration in 1958 was 315.71 ppm, the total increase is computed via the difference between them:

$Increase=414.51ppm-315.71ppm\\\\Increase=98.8ppm$

Moreover, the average increase per year is computed considering that from 1958 to 2020, 62 years have passed, therefore, such average is:

$Average\ increase/year=\frac{98.8ppm}{62 years} \\\\Average\ increase/year=1.594\frac{ppm}{year}$

Regards.

4 0

3 years ago