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2 years ago

What volume will 6.745g of neon gas occupy at standard temp and pressure?

Chemistry

2 answers:

Snezhnost [94]2 years ago

6 0

Explanation:

It is given that mass of neon is 6.745 g and molar mass of neon is 20 g/mol.

Hence, calculate the number of moles as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

Putting the given values into the above formula as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{6.745 g}{20 g/mol}$

= 0.34 mol

Also, at standard conditions value of temperature is 298 K and pressure is 1 atm.

So, putting the values into ideal gas equation and calculate the value of volume as follows.

PV = nRT

$1 atm \times V = 0.34 mol \times 0.082 L atm/mol K /times 298 K$

V = 8.31 L

Thus, we can conclude that volume of given neon gas is 8.31 L.

Thepotemich [5.8K]2 years ago

4 0

First, multiply the mass by the molar mass of neon to find out how many moles of neon there are. Then, multiply by 22.4 to find out how many liters there are.

6.745g Ne x 1 mole Ne/20 g Ne x 22.4 L/1 mole Ne = 7.5544 L

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A sample of a gas (1.50 mol) is contained in a 15.0 l cylinder. the temperature is increased from 100 °c to 150 °c. what is the

pav-90 [236]

Given:

Moles of gas, n = 1.50 moles

Volume of cylinder, V = 15.0 L

Initial temperature, T1 = 100 C = (100 + 273)K = 373 K

Final temperature, T2 = 150 C = (150+273)K = 423 K

To determine:

The pressure ratio

Explanation:

Based on ideal gas law:

PV = nRT

P= pressure; V = volume; n = moles; R = gas constant and T = temperature

under constant n and V we have:

P/T = constant

(or) P1/P2 = T1/T2 ---------------Gay Lussac's law

where P1 and P2 are the initial and final pressures respectively

substituting for T1 and T2 we get:

P1/P2 = 373/423 = 0.882

Thus, the ratio of P2/P1 = 1.13

Ans: The pressure ratio is 1.13

7 0

2 years ago

The specific heat of nickel is 0.44 J/g*⁰C. How much energy needed to change the temperature of 95.4g of nickel from 22⁰C to 32⁰

valentinak56 [21]

To determine the heat or energy needed for the process, we use the equation,
H = mcpdT
where m is the mass, cp is the specific heat and dT is the temperature difference.
H = (95.4g)(0.44 J/g°C)(32°C - 22°C)
= 419.76 J
Thus, the amount of heat that should be ABSORBED is approximately 419.76 J.

8 0

2 years ago

Read 2 more answers

Calculate the molar mass of a 2.89 g gas at 346 ml, a temperature of 28.3 degrees Celsius, and a pressure of 760 mmHg.

malfutka [58]

The molar mass of gas = 206.36 g/mol

<h3>Further explanation</h3>

In general, the gas equation can be written

$\large{\boxed{\bold{PV=nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K

T = temperature, Kelvin

mass (m)= 2.89 g

volume(V) = 346 ml = 0.346 L

T = 28.3 C + 273 = 301.3 K

P = 760 mmHg=1 atm

The molar mass (M) :

$\tt PV=\dfrac{m}{M}RT\\\\M=\dfrac{mRT}{PV}\\\\M=\dfrac{2.89\times 0.082\times 301.3}{1\times 0.346}\\\\M=206.36~g/mol$

8 0

1 year ago

(a) At what substrate concentration would an enzyme with a kcat of 30.0 s−1 and a Km of 0.0050 M operate at one-quarter of its m

Dmitrij [34]

The missing graph is in the attachment.

Answer: (a) [S] = 0.0016M

(b) Vmax = 3V; Vmax = $\frac{3V}{2}$ ; Vmax = $\frac{11V}{10}$

Explanation: Enzyme is a protein-based molecule that speed up the rate of a reaction. Enzyme Kinetics studies the reaction rates of it.

The relationship between substrate and rate of reaction is determined by the Michaelis-Menten Equation:

$V=\frac{V_{max}[S]}{K_{M}+[S]}$

in which:

V is initial velocity of reaction

Vmax is maximum rate of reaction when enzyme's active sites are saturated;

[S] is substrate concentration;

Km is measure of affinity between enzyme and its substrate;

(a) To determine concentration:

$0.25V_{max}=\frac{V_{max}[S]}{0.005+[S]}$

$0.25V_{max}(0.005+[S])=V_{max}[S]$

$0.00125+0.25[S]=[S]$

0.75[S] = 0.00125

[S] = 0.0016M

For a Km of 0.005M, substrate's concentration is 0.0016M.

(b) Still using Michaelis-Menten:

$V=\frac{V_{max}[S]}{K_{M}+[S]}$

Rearraging for Vmax:

$V_{max}=\frac{V(K_{M}+[S])}{[S]}$

(b-I) for [S] = 1/2Km

$V_{max}=\frac{V(K_{M}+0.5K_{M})}{0.5K_{M}}$

$V_{max}=\frac{V(1.5K_{M})}{0.5K_{M}}$

$V_{max}=$ 3V

(b-II) for [S] = 2Km

$V_{max}=\frac{V(K_{M}+2K_{M})}{2K_{M}}$

$V_{max}=\frac{V(3K_M)}{2K_M}$

$V_{max}=\frac{3V}{2}$

(b-III) for [S] = 10Km

$V_{max}=\frac{V(K_{M}+10K_M)}{10K_M}$

$V_{max}=\frac{V(11K_{M})}{10K_{M}}$

$V_{max}=\frac{11V}{10}$

(c) Being the affinity between enzyme and substrate, the lower Km is the less substrate is needed to reach half of maximum velocity.

Km of enzyme A is 2μM and of enzyme B is 0.5μM.

Enzyme B has lower Km than enzyme A, which means the first will need a lower concnetration of substrate to reach half of Vmax.

Analyzing each plot, notice that the red-coloured graph reaches half at a lower concentration, therefore, red-coloured plot is for enzyme B, while black-coloured plot is for enzyme A

3 0

2 years ago

A 3.96x10^-24 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000-cm cuvet; a blank solution containi

viktelen [127]

Actual question from source:-

A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000 cm cuvette. A blank had an absorbance of 0.029. The absorbance of an unknown solution of compound A was 0.375. Find the concentration of A in the unknown.

Answer:

Molar absorptivity of compound A = $1502.53\ {Ms}^{-1}$