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3 years ago

Which element has the same number of energy levels as aluminum (Al) and the same number of valence electrons as calcium (Ca)?

Chemistry

2 answers:

Zigmanuir [339]3 years ago

6 0

The correct answer is Magnesium (Mg)

scZoUnD [109]3 years ago

5 0

Answer:

magnesium

Explanation:

because i just did the test and ik this is the answer

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Which of the following is true regarding nominal and real GDP?

spin [16.1K]

d) the only thing that can increase real GDP is an increase is output

hope this helps!

5 0

3 years ago

What temperature would 3.54 moles of xenon gas need to reach to exert a pressure of 1.57 atm at a volume of 34.6 l

ira [324]

Answer:

186.9Kelvin

Explanation:

The ideal gas law equation is PV = n R T

where

P is the pressure of the gas

V is the volume it occupies

n is the number of moles of gas present in the sample

R is the universal gas constant, equal to 0.0821 atm L /mol K

T is the absolute temperature of the gas

Ensure units of the volume, pressure, and temperature of the gas correspond to R ( the universal gas constant, equal to 0.0821 atm L /mol K )

n = 3.54moles

P= 1.57

V= 34.6

T=?

PV = n R T

PV/nR = T

1.57 x 34.6/3.54 x 0.0821

54.322/0.290634= 186.908620464= T

186.9Kelvin ( approximately to 1 decimal place)

5 0

3 years ago

1.2 L sample of gas is determined to contain 0.5 moles of nitrogen. At the same temperature and pressure, how many moles of gas

hram777 [196]

A 20 L sample of the gas contains 8.3 mol N₂.

According to Avogadro’s Law, if p and T are constant

V₂/V₁ = n₂/n₁

n₂ = n₁ × V₂/V₁

___________

n₁ = 0.5 mol; V₁ = 1.2 L

n₂ = ?; V₂ = 20 L

∴n₂ = 0.5 mol × (20 L/1.2 L) = 8.3 mol

4 0

3 years ago

The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni

ValentinkaMS [17]

Answer:

" $6.7\times 10^{-4} \ atm$ " is the right answer.

Explanation:

Given:

Partial pressure of $N_2$ ,

= 0.20 atm

Partial pressure of $H_2$ ,

= 0.15 atm

$K_p = 1.5\times 10^3$ at $400^{\circ} C$

As we know,

⇒ $K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}$

By putting the values, we get

$1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}$

$pNH_3^2 = \frac{0.000675}{1.5\times 10^3}$

$=6.7\times 10^{-4} \ atm$

3 0

2 years ago

A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it r

svp [43]

Answer:

The new volume is 63583 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 5.37 * 10^4 L

The initial pressure = 0.995 atm

The initial temperature = 32.0 °C = 305.15 K

The pressure decreased to 0.720 atm

The temperature decreased to -11.7 °C = 261.45 K

Step 2: Calculate the new volume

P1*V1 / T1 = P2*V2/T2

⇒with P1 = the initial pressure = 0.995 atm

⇒with V1 = the initial volume = 5.37 *10^4 L

⇒with T1 = the initial temperature = 305.15 K

⇒with P2 = the decreased pressure = 0.720 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature : 261.45 K

(0.995 * 5.37*10^4)/305.15 = (0.720 * V2) / 261.45

V2 = 63583 L

The new volume is 63583 L

8 0

3 years ago