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hram777 [196]
2 years ago
10

What are two fungal skin infections that humans can contract?

Chemistry
1 answer:
Natali5045456 [20]2 years ago
5 0
Ringworm and jock itch
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Which statement below is the BEST explanation for the differences
IgorC [24]

Answer:

A

Explanation:

Electron geometry describes the arrangement of electron groups. Molecular geometry describes the arrangement of atoms, excluding lone pairs.

7 0
3 years ago
effect. RbClO(s) → Rb+(aq) + ClO−(aq) HClO(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + ClO−(aq) The degree of dissociatio
Lemur [1.5K]

Explanation:

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of strontium sulfate and sodium sulfate follows the equation:

RbClO(s)\rightarrow Rb^{+}(aq.)+ClO^{-}(aq.)

HClO(aq)+H_2O\rightleftharpoons H_3O^+(aq.)+ClO^{-}(aq.)

According to Le-Chateliers principle: If there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, hypochlorite ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of hydrogen hypochlorite.

Thus, the addition hypochlorite ions will shift the equilibrium in the left direction.

The dissociation of hydrogen hypochlorite is suppressed due to the common ion effect.

8 0
3 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
To evaluate R in an Ideal gas Law (PV=nRT), the Volume of a gas is measured in ____ , while the temperature of a gas is measured
Vika [28.1K]
L (litter)
K (kelvin)
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Una nueva marca de bebidas disuelve frutosa (C6H12O6) en etanol (C2H5OH) como endulzante, con el fin de cumplir con los controle
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Explanation:

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