Ask question

Ask question

All categories

3 years ago

10

Sodium an alkali metal and chlorine a halogen are both in period 3?of the periodic table. Which element has a higher ionization

energy? Explain

1 answer:

V125BC [204]3 years ago

7 0

Answer:

chlorine

Explanation:

Because in periodic table across the period from left to right, ionization energy increases due to increase in the nuclear charge as more and more electrons are added to the same shell.

You might be interested in

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

8090 [49]

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is $P_A$ and vapour pressure of component B is $P_B$ .

$P^0_A$ = Vapour pressure of component A in pure form

$P^0_B$ = Vapour pressure of component B in pure form

$x_A$ =Mole fraction of component A

$x_B=$ =Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

$P_A >P^0_A\cdot x_A$ , $P_B>P^0_B\cdot x_B$

Therefore, $P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B$

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

6 0

3 years ago

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

3 years ago

Which components of the apparatus affect in motion? If you answer this and it has non to do with the question imma report u!!!!

Alenkinab [10]

Answer:

Explanation:

An internal force acts between elements of the system. Only external forces affect the motion of a system, according to Newton's first law. Newton's second law states that a net force on an object is responsible for its ... Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological

3 0

3 years ago

Given the reactant side of the total ionic equation for the neutralization reaction of lithium hydroxide (LiOH) with hydrochlori

Sunny_sXe [5.5K]

<span>Answer:
</span><span>
</span><span>
</span><span>Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)</span><span />

<span>Explanation:
</span>

<span>1) Combine the cation Li⁺ (aq) with the anion Cl- (aq) to form LiCl(s).
</span>
<span>LiCl is a solid soluble substance, a typical ionic compound. So, it will reamain as separate ions in the product side: Li⁺ + CL⁻</span>

<span>2) Combine the anion OH⁻ with the cation H⁺ to form H₂O(l).
</span>
<span>Since, the ionization of H₂O is low, it will remain as liquid in the product side: H₂O(l)</span>

<span>3) Finally, you can wirte the total ionic equation:
</span>
Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)

8 0

3 years ago

What is the coefficient for H2O when the equation ? Ca(OH)2(aq) + ? H3PO4(aq) → ? Ca3(PO4)2(s) + ? H2O(ℓ) is balanced using the

kari74 [83]

Answer:

Coefficient of $H_{2}O$ in the balanced equation with smallest possible integer is 6.

Explanation:

Unbalanced equation: $Ca(OH)_{2}(aq.)+H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)$

Balance Ca: $3Ca(OH)_{2}(aq.)+H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)$

Balance $PO_{4}$ : $3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)$

Balance H and O: $3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+6H_{2}O(l)$

Balanced equation: $3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+6H_{2}O(l)$

So coefficient of $H_{2}O$ in the balanced equation with smallest possible integer is 6.

7 0

3 years ago