In N, +0, N,O,, nitrogen is reduced A True B. False

An isotope with a mass number of 193 has 116 neutrons. What is the atomic number of this

Zinaida [17]

Answer:

Atomic number of this isotope = 77

Explanation:

Given that,

Mass number = 193

No of neutrons = 116

We need to find the atomic no of this isotope.

We know that,

Atomic mass = No of protons + No. of neutrons

Also, atomic no = no of protons

So,

Atomic mass = atomic no + No. of neutrons

⇒ Atomic no = Atomic mass - no of neutrons

Atomic no = 193 - 116

Atomic no = 77

Hence, 77 is the atomic no of the isotope.

6 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

2 years ago

When PCl5 solidifies it forms PCl4+ cations and PCl6– anions. According to valence bond theory, what hybrid orbitals are used by

Vlad1618 [11]

Answer:

sp³

Explanation:

Number of hybrid orbitals = ( V + S - C + A ) / 2

Where

H is the number of hybrid orbitals

V is the valence electrons of the central atom = 5

S is the number of single valency atoms = 4

C is the number of cations = 1

A is the number of anions = 0

For PCl₄⁺

Applying the values, we get:

H = ( 5+4-1+0) / 2

= 4

This corresponds to sp³ hybridization.

5 0

3 years ago

Read 2 more answers

CH3CH2OH + 302 → 2C02 + 3H2O

Colt1911 [192]

Answer:

Oxygen Gas

Explanation:

The balanced equation shows us the reactant ratio of the reaction.

This means that for every one mole of CH3CH2OH, we need 3 moles of O2 to react with it. Because we need more O2, (3x as much) than ethanol and we have the same given amount (1 mole of each), the oxygen will be the limiting reagent. (1 mole of oxygen would only require 1/3 moles of ethanol to react).

Hope this helped!

7 0

2 years ago

Calculate the empirical formula for DCBN, given that the percent composition is 48.9%C, 1.76%H, 41.2%Cl, and 8.15%N.

sweet [91]

Answer:

empirical formula is C7H3NCl2

Explanation:

too much work too explain and im lazy

5 0

3 years ago

In N, +0,N,O,, nitrogen is reducedA TrueB. False​

In N, +0,N,O,, nitrogen is reducedA TrueB. False