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3 years ago

how does the energy required to sublime a substance compare to the energy required to melt the substance

1 answer:

3 0

The energy required to sublime (solid to gas) a substance at 1 ATM pressure is greater than the energy required to melt (solid to liquid) a substance. When you compare the energies in varying pressures, however, this trend is not always the case.

any 'general phase diagram', you can see that under the triple point, when all phases are in equilibrium, have solid and gas meeting under a certain pressure. In a vacuum, it would require less energy to sublime than to melt.

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Which of the following elements has electrons in the 3d sublevel?

ch4aika [34]

Answer:

fe is the answer

Explanation:

3 0

2 years ago

MnO2 + 4HCl reacts to form MnCl2 + CL2 + 2H2O If .86 mol of MnO2 and 48.2 g of HCl react, which reactant is the limiting reagent

Katen [24]

HCl:

m=48,2g
M=36,5g/mol

n = m/M = 48,2g / 36,5g/mol = 1,32mol

1mol : 4mol
MnO₂ + 4HCl ⇒ MnCl₂ + Cl₂ + 2H₂O
0,86mol : 1,32mol
limiting reagent
0,33 will react

HCl is limiting reagent.

6 0

3 years ago

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

2 years ago

Ammonia and oxygen react to form nitrogen monoxide and water, like this:

olchik [2.2K]

4NH3+5O2 <=>4NO + 6H2O

Using the definition of Kp, we have
Kp=(Pno^4*Ph2o^6)/(Pnh3^4*Po2^5)
where Pno=partial pressure of NO, etc.

The numerical value for a given temperature can be evaluated when the actual partial pressures are known.

6 0

3 years ago

What’s the answer pls hrlp

arsen [322]

Answer:

The answer is A

Explanation:

5 0

3 years ago