What element has 2 electrons in its outermost energy level and is in the 2nd period on periodic table?

.an open flask contains 0.200 mol of air. atmospheric pressure is 745 mmhg and room temperature is 68˚f. how many moles are pres

aleksklad [387]

0.219 moles moles are present in the flask when the pressure is 1.10 atm and the temperature is 33˚c.

What is ideal gas constant ?

The ideal gas constant is calculated to be 8.314J/K⋅ mol when the pressure is in kPa.
The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.
The combined gas law relates pressure, volume, and temperature of a gas.

We simple use this formula-

The basic formula is PV = nRT where. P = Pressure in atmospheres (atm) V = Volume in Liters (L) n = of moles (mol) R = the Ideal Gas Law Constant.

68F = 298.15K

V = nRT/P = 0.2 * 0.08206 * 298.15K / (745/760) = 4.992Liters

n = PV/RT = 1.1atm*4.992L/(0.08206Latm/molK * 306K)

n = 0.219 moles

Therefore, 0.219 moles moles are present in the flask when the pressure is 1.10 atm and the temperature is 33˚c.

Learn more about ideal gas constant

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4 0

2 years ago

Is 2 KNO₃ + H₂CO₃ → K₂CO₃ + 4 HNO₃ a balanced equation

krok68 [10]

Hey there!

The elements in this equation are K, N, O, H, and C.

Let's count how many of each are on each side to see if it is balanced.

K: 2 on the left, 2 on the right.

N: 2 on the left, 4 on the right.

O: 9 on the left, 6 on the right.

H: 2 on the left, 4 on the right.

C: 1 on the left, 1 on the right.

Notice that there are different amounts of N, O, and H on the left side and the right side.

This means that the equation is not balanced.

Hope this helps!

6 0

3 years ago

A sample of 211 g of iron (III) bromide is reacted with

Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

$\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714$

Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

$\tt n=\dfrac{186}{78.0452}=2.38$

Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :

$\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793$

So FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

$\tt \dfrac{6}{3}\times 0.714=1.428$

6 0

3 years ago

What is the product of 1.65 and 1.05 x 10-27

lilavasa [31]

Answer:

-9.675

Explanation:

correct me if im wrong.!!

3 0

3 years ago

Read 2 more answers

Under standard-state conditions, which of the following half-reactions occurs at the cathode during the electrolysis of aqueous

Slav-nsk [51]

Answer:

The following reaction will occur at cathode:

$Ni^{+2}(aq)+2e--->Ni(s)$

Explanation:

The two half reaction during electrolysis of aqueous nickel sulfate will be

a) anode reaction :

Water will undergo oxidation and will evolve oxygen gas at anode as shown in the given reaction:

$2H_{2}O(l) ----> O_{2}(g) + 4H^{+}(aq) + 4e$

b) Cathode reaction: The reduction of Nickel ion will occur by gain of two electrons as shown in the given equation:

$Ni^{+2}(aq)+2e--->Ni(s)$

Thus the overall reaction will be:

$2H_{2}O(l) + 2Ni^{+2}----> O_{2}(g) + 4H^{+}(aq) + 2Ni(s)$

6 0

3 years ago