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3 years ago

12 points

Chemistry

1 answer:

Ad libitum [116K]3 years ago

8 0

Answer:

C₅H₈O₂

Explanation:

methyl methacrylate = 100 amu

6.91g CO₂ = 0.157 moles

2.26g H₂O = 0.125 moles

0.157 ÷ 0.125 = 1.256

{(CO₂)₁.₂₅₆ + (H₂O)₁} × 4 = (CO₂)₅ + (H₂O)₄

C₅H₈O?

C₅ = 60.05 amu H₈ = 8.064 amu

60.05 + 8.064 = 68.114 amu

100 amu - 68.114 amu = 31.886 amu

O = 16 amu

O = 2

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3 years ago

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A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin

lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, $3.0\times 10^1g$

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron metal = $0.45J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

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$m_2$ = mass of water = ?

$T_f$ = final temperature of mixture = $59.2^oC$

$T_1$ = initial temperature of iron metal = $21.9^oC$

$T_2$ = initial temperature of water = $63.5^oC$

Now put all the given values in the above formula, we get

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3 years ago

PLEASE HELP!!! NEED TO PASS THIS TO THE FIRST SEMESTER!

Cloud [144]

Answer:

Reaction 1 is balanced but 2 is not balanced , the balance equation are :

1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

2. $CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + H_{2}O(aq)$

Explanation:

Balanced Equations : These are the equation which follows the law of conservation of mass .

The total number of atoms present in reactant is equal to total number of atoms present in product.

1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

This is acid - base type reaction where

$CH_{3}COOH(aq)$ act as Acid

$NaHCO_{3}(aq)$ act as weak base

Reactant : $CH_{3}COOH(aq)$ , $NaHCO_{3}$

Number of atoms of :

C = 2 ( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 2 + 1

= 3

H = 4( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 4 + 1

O = 2( $CH_{3}COOH(aq)$ ) + 3 ( $NaHCO_{3}$ )

= 5

Na = 1 ( $NaHCO_{3}$ )

= 1

Product : $CO_{2}(g)$ , $H_{2}O(l)$ , $CH_{3}COONa(aq)$

Number of atoms :

C = 1( $CO_{2}(g)$ ) + 2( $CH_{3}COONa(aq)$ )

= 1 + 2

= 3

H = 2( $H_{2}O(l)$ ) + 3( $CH_{3}COONa(aq)$ )

= 2 + 3

= 5

O = 1( $H_{2}O(l)$ ) + 2( $CH_{3}COONa(aq)$ )

+2( $CO_{2}(g)$

= 1 + 2 + 2

= 5

Na = 1( $CH_{3}COONa(aq)$

= 1

Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction

2. $CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + CO_{2}(g)$

This is double displacement reaction .

Check the balancing in both reactant and products should be :

Na = 2

H = 2

Ca = 1

C = 2

O = 6

Cl = 2

7 0

3 years ago