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Natali5045456 [20]

3 years ago

11

Find the source frequency ω for which the equivalent impedance seen by the voltage source is purely real (i.e., the imaginary pa

rt of the equivalent impedance is j0).

1 answer:

Leto [7]3 years ago

8 0

I just took this quiz and the answer is d

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Give explanations for the large drops in melting point from C to Si and from Ge to Sn.

____ [38]

As we move down the group, the metallic bond becomes more stable and the formation of forming covalent bond decreases down the group due to the large size of elements.

Covalent and metallic bonding leads to higher melting points. Due to a decrease in attractive forces from carbon to lead there is a drop in melting point.

Carbon forms large covalent molecules than silicon and hence has a higher melting point than silicon.

Similarly, Ge also forms a large number of covalent bonds and has a smaller size as compared to that of Sn. Hence melting point decreases from Ge to Sn.

The order will be C>Si>Ge>Pb>Sn.

To learn more about the covalent bond, visit: brainly.com/question/10777799

#SPJ4

3 0

1 year ago

Which reaction results in only a physical change?

Advocard [28]

D will because water just turns from liquid to a gas by adding heat. That is a physical change. The other equations are chemical changes.

4 0

4 years ago

What type of reaction occurs between aqueous ammonia solution and an acid?

gizmo_the_mogwai [7]

Neutralization reaction.

8 0

3 years ago

Read 2 more answers

Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th

stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH= $2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}$

Knowing:

ΔH= 75.5 kJ/mol
$H_{NO}$ = 90.25 kJ/mol
$H_{Cl_{2} }$ = 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
$H_{NOCl}$ =?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 - $H_{NOCl}$

Solving

- $H_{NOCl}$ =75.5 kJ/mol - 2*90.25 kJ/mol

- $H_{NOCl}$ =-105 kJ/mol

$H_{NOCl}$ =105 kJ/mol

<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>

8 0

3 years ago

In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to

labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of $NH_3$ = 100 g

Molar mass of $NH_3$ = 27 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate moles of $NH_3$ .

$\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles$

The given balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the given reaction, we conclude that

2 moles of $NH_3$ produced from 1 mole of $N_2$

3.7 moles of $NH_3$ produced from $\frac{1mole}{2mole}\times 3.7mole=1.85moles$ of $N_2$

Now we have to calculate the mass of $N_2$ .

Mass of $N_2$ = Moles of $N_2$ × Molar mass of $N_2$

Mass of $N_2$ = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0

3 years ago