Balance the following equations:
i. H2SO4 + ….NaHCO3 → Na2SO4 + ….H2O + ……CO2
ii. H2 + I2 → ….HI
iii. …..NaOH + H2SO4 → Na2SO4 + …..H2O
iv. ……FeSO4 → Fe2O3 + SO2 + SO3
v. C3H6O2 + …..O2 → ….CO2 + …..H2O
b) Calculate the molecular masses of the following:
i. H2SO4

ii. CH4

iii. NH3

c) 56g of iron reacts with 71g of chlorine. How many grams of iron react with 35.5g of chlorine? ​

Answers

Answer 1
Hey There!_____________________________________Answer:_____________________________________Balancing the following equations:

(I) [tex]H_2SO_4+ 2NaHCO_3--> Na_2SO_4 + 2 H_2O + 2CO_2[/tex]

(II) [tex]H_2+I_2 --> 2HI[/tex]

(III) [tex]2NaOH+H_2SO_4--> Na_2SO_4+ 2H_2O[/tex]

(IV) [tex]2FeSO_4--> Fe_2O_3+SO_2+SO_3[/tex]

(V) [tex]2C_3H_6O_2+7O_2-->6CO_2+6H_2O[/tex]

_____________________________________Molecular Masses of the following:

------------------------------------------------------------------------------------------------------------

(I) [tex]H_2SO_4[/tex]

(ANS) [tex]98 \frac{grams}{mol}[/tex]

__________________________________________________________

Solution:

Molecular mass of,

H = 1

S = 32

O = 16

Now,

There are 2 atoms of Hydrogen, 1 atoms of Sulfur and 4 atoms of Oxygen present, thus,

[tex]H_2SO_4 = (1x2)+32+(16x4)[/tex]

[tex]H_2SO_4 = 98 \frac{grams}{mol}[/tex]

_____________________________________

(II) [tex]CH_4[/tex]

(ANS) [tex]16 \frac{gram}{mol}[/tex]

_________________________________________________________

Solution:

Molecular Mass,

Carbon = 12

Hydrogen = 1

Now

There are 1 atom of Oxygen and 4 atoms of Hydrogen present, thus,

[tex]CH_4 = 12 + (1x4)[/tex]

[tex]CH_4 = 16\frac{grams}{mol}[/tex]

_____________________________________

(III) [tex]NH_3[/tex]

(ANS) [tex]17 \frac{grams}{mol}[/tex]

__________________________________________________________

Solution:

Molecular Mass,

Nitrogen = 14 grams

Hydrogen = 1

Now,

There are 1 atoms of Nitrogen and 3 atoms of Hydrogen present, thus,

[tex]NH_3 = 14 + (1x3)[/tex]

[tex]NH_3 = 17 \frac{grams}{mol}[/tex]

_____________________________________How many Grams of Iron(fe):[tex]Fe + Cl_2 --> FeCl_2[/tex]

56     71

x       35.5

_____________________________________

[tex]X = \frac{56 x 35.5}{71}[/tex]

Simplify the equation

X = 28 grams

__________________________________________________________

Answer:

28 grams of Iron(fe) will react with chlorine.

_____________________________________Best Regards,'Borz'

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Answer:

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Explanation:

Answer:

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Explanation:

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Answer:

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Answer:

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Answer:

A

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Explanation:

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Answers

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Further explanation

Given

A half life of 2.5 days ⇒ t1/2 = 2.5 days

initial sample ⇒ 25 g

duration of decay ⇒ 10 days

Required

sample left after 10 days

Analysis

Use decay Formula :

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Solution

[tex]\tt Nt=25.\dfrac{1}{2}^{10/2.5}\\\\Nt=25.\dfrac{1}{2}^4\\\\Nt=\boxed{\bold{1.563~g}}[/tex]

Paraphrase

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Answer:

heterogeneous          

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Answers

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Answer:

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