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3 years ago

6

Why do magnesium, phosphorus, and zinc exhibit slightly higher first ionization energies than the general trend within each of t

heir periods?

1 answer:

Lapatulllka [165]3 years ago

8 0

Because they are farther across the periodic table<span />

You might be interested in

If 13.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

Paul [167]

6.349 g mass of anhydrous magnesium sulfate will remain.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Molar mass MgSO₄.7 H₂O = 246.52 g/mol

$Moles =\frac{mass}{molar \;mass}$

$Moles =\frac{13.0 g}{246.52}$

0.0527 moles

Molar mass MgSO₄ = 120.4 g/mol

Mass of anhydrous magnesium sulfate :

( 0.0527 x 120.4 ) => 6.349 g

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

4 0

2 years ago

Is a molecule the same as a compound ?

Alenkasestr [34]

No a molecule is 2 different atoms bond

5 0

3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

3. What do you notice about the temperature of the substance in the test tube after it is placed in the beaker of water?

34kurt

Answer:

C: The temperature of the substance increases as it sits in the beaker of water

Explanation:

This question was taken from a video where an attempt was made to investigate the changes in temperature when a substance undergoes change from it's solid phase to its liquid phase.

To do this, as seen in the video online, it shows a solid substance in a test tube being placed in a beaker of water.

From observation, the water in the beaker has a warmer temperature than the solid substance present in the test tube and this in turn makes the test tube gradually increase in temperature.

Thus, the solid substance will as well increase increase in temperature when it is placed in the beaker of water.

3 0

3 years ago

What is the molecular formula of the hydrocarbon whose molar mass is 536 g/mol and contains 89.55 w/w% carbon?

leonid [27]

Given the percentage composition of HC as C → 81.82 % and H → 18.18 %

So the ratio of number if atoms of C and H in its molecule can will be:

C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8

So the Empirical Formula of hydrocarbon is:

C 3 H 8

As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol

Now let Molecular formula of the HC be ( C 3 H 8 ) n

Using molar mass of C and H the molar mass of the HC from its molecular formula is:

( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1

Hence the molecular formula of HC is C 3 H 8

Does that help?

7 0

3 years ago