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Rom4ik [11]
4 years ago
11

Using the information from STP or SATP conditions determine the value of the ideal gas constant.

Chemistry
1 answer:
Dovator [93]4 years ago
4 0

Answer:

0.0821 atm.L/Kmol

Explanation:

At stp, the values temperature, pressure and volume is given below:

Pressure (P) = 1 atm

Temperature (T) = 273 K

Volume (V) = 22.4 L

At stp, 1 mole of a gas occupy 22.4L.

Number of mole (n) = 1 mole

Gas constant (R) =?

The ideal gas equation is given below:

PV = nRT.

With the above equation, the gas constant R can be obtained as follow:

1 atm x 22.4L = 1 mole x R x 273K

Divide both side by (1 mole x 273 K)

R = (1 atm x 22.4L) / (1 mole x 273 K)

R = 0.0821 atm.L/Kmol

Therefore, the gas constant is 0.0821 atm.L/Kmol

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Significant Figures For 7.06 × 10^5 ÷ 5.3 × 10^-2
Kay [80]

Answer:

\boxed{\text{two}}

Explanation:

In multiplication  and division problems, your answer can have no more significant figures than the number with the fewest significant figures.

\dfrac{7.06 \times 10^{5}}{5.3 \times 10^{-2}}= 1.332 075 472 \times 10^{7} (by my calculator)  

There are three significant figures in 7.06 and two in 2.3.

You must round to \boxed{\textbf{two}} significant figures and report the answer as 1.3 × 1.0⁷.

4 0
3 years ago
A substance undergoes a change. Which of the following indicates that the change
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Answer:

change of color

Explanation:

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3 years ago
Ocean currents - Surface ocean currents can occur on local and global scales and are typically wind-driven, resulting in both ho
Aleksandr [31]
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4 0
3 years ago
Carbon-14 has a half-life of 5,730 years. If scientists discover a mummified Cro-Magnon specimen, they will first want to determ
ryzh [129]

Answer:

the specimen has 11460 years old

Explanation:

if the live sample has as initial amount of Yo C14, the dead sample will have 0.25Yo of C14.

the rate of decay of radiactive matter over time is

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∴ Y(t) = 0.25Yo;  T = 5730

⇒ 0.25Yo = Yo * e (( - t * Ln2 ) / 5730 )

⇒ 0.25 = e (( - t * Ln2 ) / 5730 )

⇒ Ln(0.25) = ( -  t * Ln2 ) / 5730

⇒ - 7943.466 = - t * Ln2

⇒ 7943.466 / Ln2 = t

⇒ t = 11460 year

4 0
3 years ago
How many grams are in 2.3 x 10^-4 moles of Ca3(PO4)2
OLEGan [10]
Molar mass 

Ca₃(PO₄)₂ = 310 g/mol

1 mole -------------------> 310 g
2.3x10⁻⁴ mole ---------> ?

m = 2.3x10⁻⁴ * 310 / 1

m = 0.0713 g 

hope this helps!
3 0
3 years ago
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