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pickupchik [31]
3 years ago
12

Explain why vanadium (radius=134 pm) and copper (radius=128 pm) have nearly identical atomic radii, even though the atomic numbe

r of copper is about 25% higher than that of vanadium.
What would you predict about the relative densities of these two metals? Look up the densities in a reference book, periodic table, or on the Web.
Are your predictions correct?
Chemistry
1 answer:
Viktor [21]3 years ago
7 0
<span>There are few main factors affecting the atomic radii, the outermost electrons and the protons in the nucleus and also the shielding of the internal electrons. I would speculate that the difference in radii is given by the electron clouds since the electrons difference in these two elements is in the d orbital and both has at least 1 electron in the 4s (this 4s electron is the outermost electron in all the transition metals of this period). The atomic radio will be mostly dependent of these 4s electrons than in the d electrons. Besides that, you can see that increasing the atomic number will increase the number of protons in the nucleus decreasing the ratio of the atoms along a period. The Cu is an exception and will accommodate one of the 4s electrons in the p orbital.

</span><span>Regarding the density you can find the density of Cu = 8.96g/cm3 and vanadium = 6.0g/cm3. This also correlates with the idea that if these two atoms have similar volume and one has more mass (more protons;  density is the relationship between m/V), then a bigger mass for a similar volume will result in a bigger density.</span>
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3. Hydrogen reacts with nitrogen to produce ammonia according to the equation:
lys-0071 [83]

Mass of ammonia produced : 121.38 g

<h3>Further explanation</h3>

Given

Reaction

3H₂(g) + N₂(g) ⇒ 2NH₃(g)

100g of N₂

Required

Ammonia produced

Solution

mol of N₂ :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{100}{28}\\\\mol=3.57

From the equation, mol ratio of N₂ and NH₃ = 1 : 2, so mol NH₃ :

\tt \dfrac{2}{1}\times 3.57=7.14~moles

mass of NH₃(MW=17 g/mol) :

\tt mass=mol\times MW\\\\mass=7.14\times 17\\\\mass=121.38~g

8 0
2 years ago
How to balance the chemical equation: <br> CO2+H2O--&gt;C6H12O6
Andru [333]

Answer:

6H₂O + 6CO₂ + energy  →   C₆H₁₂O₆ + 6O₂

Explanation:

The given reaction represent the formation of glucose so it is photosynthesis reaction.

Photosynthesis:

It is the process in which in the presence of sun light and chlorophyll by using carbon dioxide and water plants produce the oxygen and glucose.

Carbon dioxide + water + energy →   glucose + oxygen

water is supplied through the roots, carbon dioxide collected through stomata and sun light is capture  by chloroplast.

Chemical equation:

6H₂O + 6CO₂ + energy  →   C₆H₁₂O₆ + 6O₂

it is known from balanced chemical equation that 6 moles of carbon dioxide react with the six moles of water and created one mole of glucose and six mole of oxygen.

4 0
3 years ago
Oxygen and hydrogen are both elements that are found as gases at room temperature. When oxygen combines with hydrogen, they prod
Tom [10]

Answer:

A

Explanation:

has properties that are different from the original substances.

5 0
2 years ago
3.
sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5.  V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a)  ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0
3 years ago
The frequency of an x-ray wave is 3.0 x 1012 MHz. Its wave speed is 3.0x 108 m/s. Calculate the wavelength of the x-ray wave bel
elena55 [62]

Answer:

λ = 1*10⁻¹⁰m

Explanation:

Frequency (f) = 3.0*10¹²MHz = 3.0*10¹⁸Hz

Speed (v) = 3.0*10⁸m/s

Speed (v) of a wave = frequency (f) * wavelength (λ)

V = fλ

Solve for λ,

λ = v / f

λ = 3.0*10⁸ / 3.0*10¹⁸

λ = 1*10⁻¹⁰m

λ = 0.

6 0
2 years ago
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