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2 years ago

State the phenomenon exhibited by the two atoms

Chemistry

1 answer:

Volgvan2 years ago

6 0

Answer:

METALS

Lose their valence electrons easily/ ionic by electron loss.

NOMETAL

Gain or share valence electrons easily/ ionic by electron grain.

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Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you

Alja [10]

Answer: The molality of the solution is 0.11 m

Explanation:

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

$0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g$

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (methanol) = 0.135 g

$M_{solute}$ = Molar mass of solute (methanol) = 32 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

$\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m$

Hence, the molality of the solution is 0.11 m

6 0

3 years ago

Equation Given : Al^(3+) + Na3PO4 ==> 3Na^+ + AlPO4

Helga [31]

1 mols of Aluminium ion forms 1 mol aluminium phosphate

Molar mass of AlPO_4

27+31+16(4)
58+48
106u

Moles of AlPO_4

61µg/106
0.000061/106
5.75×10^{-7}
57.5µmol

Moles of Al3+=57.5µmol

3 0

2 years ago

To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w

Elis [28]

Answer: The number of moles of weak acid is $4.24\times 10^{-3}$ moles.

Explanation:

To calculate the moles of KOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}$

We are given:

Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

$0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol$

The chemical reaction of weak monoprotic acid and KOH follows the equation:

$HA+KOH\rightarrow KA+H_2O$

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, $4.24\times 10^{-3}mol$ of KOH will react with = $\frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol$ of weak monoprotic acid.

Hence, the number of moles of weak acid is $4.24\times 10^{-3}$ moles.

6 0

3 years ago

Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this

maw [93]

Answer: d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

$2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H=+1411.1kJ$

Reversing the reaction, changes the sign of $\Delta H$

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$

$\Delta H=-1411.1kJ$

On multiplying the reaction by $\frac{1}{2}$ , enthalpy gets half:

$0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)$ $\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol$

Thus the enthalpy change for the given reaction is -705.55kJ

7 0

3 years ago

What must be true when work is done?

klio [65]

In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force. ... For the special case of a constant force, the work may be calculated by multiplying the distance times the component of force which acts in the direction of motion.

7 0

3 years ago

State the phenomenon exhibited by the two atoms​

State the phenomenon exhibited by the two atoms