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3 years ago

Compare and contrast the properties of halogens & alkali metals

Chemistry

1 answer:

zhuklara [117]3 years ago

4 0

Answer:

Physical Properties

Alkali metals are good conductors of heat and electricity while halogens are poor conductors of heat and electricity.

Only specific to Alkali metals: They are soft and can be cut by knife.

Chemical Properties

Alkali metals have 1 valence electron while halogens have 7 valence electrons.

Alkali metals react with non-metals to form ionic compounds while halogens react with non-metals to form covalent compounds.

You can also look at the trends ( melting/boiling point, reactivity) down their group.

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Calculate the energy (in kJ) required to heat 10.1 g of liquid water from 55 oC to 100 oC and change it to steam at 100 oC. The

Maksim231197 [3]

Answer:

$\large\boxed{\large\boxed{24.6kJ}}$

Explanation:

1. Energy to heat the liquid water from 55ºC to 100ºC

$Q=m\times C\times \Delta T$

m = 10.1g

C = 4.18g/JºC

ΔT = 100ºC - 55ºC = 45ºC

$Q=10.1g\times 4.18J/g\ºC\times 45\ºC=1,899.81J$

2. Energy to change the liquid to steam at 100ºC

$L=\lambda \times n$

λ = 40.6kJ/mol

n = 10.1g / 18.015g/mol = 0.5606mol

$L=40.6kJ/mol\times 0.5604mol=22.76214kJ=22,762.14J$

3. Total energy

$1,899.81J+22,762.14J=24,661.95J\approx24,662J\approx24.6kJ$

7 0

4 years ago

A solution is prepared by adding 16 g of CH3OH (molar mass 32 g) to 90 g of H2O (molar mass 18 g). The mole fraction of CH3OH in

andreyandreev [35.5K]

Answer:

The mole fraction of CH3OH in this solution is closest to 0.1.

Explanation:

The mole fraction of CH3OH can be calculated by dividing the number of mole of CH3OH by the sum of all moles present in the solution. In our example, we have 16 g CH3OH and 90 g H2O. Let´s see how many mols of each constituent we have:

1 mol CH3OH = 32 g ⇒ 16 g CH3OH = 0.5 mol.

1 mol H2O = 18 g ⇒ 90 g H20 = 5 mol

Then, the mole fraction is:

X = mol CH3OH / (mol H2O + mol CH3OH)

X= 0.5 mol / (5 mol + 0.5 mol) ≅ 0.1

6 0

3 years ago

: Starting with 0.3500 mol CO(g) and 0.05500 mol COCl2(g) in a 3.050-L flask at 668 K, how many moles of Cl2(g) will be present

Mrac [35]

Answer:

The number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

Explanation:

The reaction is:

CO(g) + Cl₂(g) ⇄ COCl₂(g)

The equilibrium constant of the above reaction is:

K = 1.2x10³

To find the moles of Cl₂ present at equilibrium, let's evaluate the reverse reaction:

COCl₂(g) ⇄ CO(g) + Cl₂(g)

The equilibrium constant for the reverse reaction is:

$K_{r} = \frac{1}{1.2 \cdot 10^{3}} = 8.3 \cdot 10^{-4}$

Now, we need to calculate the concentration of CO and COCl₂:

$C_{CO} = \frac{\eta_{CO}}{V} = \frac{0.3500 moles}{3.050 L} = 0.115 M$

$C_{COCl_{2}} = \frac{\eta_{COCl_{2}}}{V} = \frac{0.05500 moles}{3.050 L} = 0.018 M$

Now, from the reaction we have:

COCl₂(g) ⇄ CO(g) + Cl₂(g)

0.018 - x 0.115+x x

The concentration of Cl₂ is:

$K_{r} = \frac{[CO][Cl_{2}]}{[COCl_{2}]}$

$8.3 \cdot 10^{-4} = \frac{(0.115 + x)(x)}{0.018 - x}$

$8.3 \cdot 10^{-4}*(0.018 - x) - (0.115 + x)(x) = 0$

By solving the above equation for x we have:

x = 1.29x10⁻⁴ M = [Cl₂]

Finally, the number of moles of Cl₂ present at equilibrium is:

$\eta_{Cl_{2}} = C_{Cl_{2}}*V = 1.29 \cdot 10^{-4} mol/L*3.050 L = 3.94 \cdot 10^{-4} moles$

Therefore, the number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

I hope it helps you!

8 0

3 years ago

A sample of water with a mass of 622.7g undergoes a temperature change from 90.98°C

Dovator [93]

Answer:

C) Q = 14888.5 J

Explanation:

Q = mCΔT

∴ m H2O = 622.7 g

∴ C H2O = 4.18 J/g°C

∴ ΔT= 96.7°C - 90.98°C = 5.72°C

⇒ Q = (622.7 g)(4.18 J/g°C)(5.72°C)

⇒ Q = 14888.508 J

6 0

4 years ago

2. What are two types of local winds?

netineya [11]

Answer:

The two main types of local winds are:

1. Sea Breezes.

2. Land Breezes.

4 0

3 years ago

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