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Nataly [62]
3 years ago
11

1 which equation demonstrates the law of conservation of mass?

Chemistry
1 answer:
Mrac [35]3 years ago
5 0
Every balanced chemical reaction demonstrates the law of conservation of mass where the reactants have the same number of atoms as the products. For example, the production of ammonia gas.
N2 (g) + 3H2 (g) -> 2NH3 (g) you can see that there are two nitrogen atoms and six hydrogen atoms on both sides of the equation.
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Which option is an example of a chemical property?
Alona [7]
Answer:
The ability to decompose
Explanation:
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Density and color are both physical properties.
Even though melting point may seem like a chemical property, when something melts, only the physical state changes, and the chemical structure does not change, and therefore, is a physical property.
The ability to decompose is a chemical property. When something decomposes, the chemical structure of many molecules change, and therefore, is classified as a chemical property.
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6 Fe2+ (aq) + Cr2O72− (aq) + 14 H+ (aq) → 6 Fe3+ (aq) + 2 Cr3+ (aq) + 7 H2O (aq) If the titration of 23 mL of an iron(II) soluti
ratelena [41]

Answer:

1.047 M

Explanation:

The given reaction:

6Fe^{2+}_{(aq)}+Cr_2O_7^{2-}_{(aq)}+14H^+_{(aq)}\rightarrow 6Fe^{3+}_{(aq)}+2Cr^{3+}_{(aq)}+7H_2O_{(aq)}

For dichromate :

Molarity = 0.254 M

Volume = 15.8 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.8 ×10⁻³ L

Thus, moles of dichromate :

Moles=0.254 \times {15.8\times 10^{-3}}\ moles

Moles of dichromate = 0.0040132 moles

1 mole of dichromate react with 6 moles of iron(II) solution

Thus,

0.0040132 moles of dichromate react with 6 × 0.0040132 moles of iron(II) solution

Moles of iron(II) solution = 0.02408 moles

Volume = 23 mL = 0.023 L

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

<u>Molarity = 0.02408 / 0.023 = 1.047 M</u>

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3 years ago
True or false all countries are legally bound to survey and monitor for disease within their bodies
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Answer:

False..........

7 0
3 years ago
Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen
olga55 [171]

Answer: Rate law=k[A]^1[B]^2, order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is 3L^2mol^{-2}s^{-1}

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Rate=k[A]^x[B]^y

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: 1.2\times 10^{-2}=k[0.10]^x[0.20]^y    (1)

From trial 2: 4.8\times 10^{-2}=k[0.10]^x[0.40]^y    (2)

Dividing 2 by 1 :\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}

4=2^y,2^2=2^y therefore y=2.

b) From trial 2: 4.8\times 10^{-2}=k[0.10]^x[0.40]^y    (3)

From trial 3: 9.6\times 10^{-2}=k[0.20]^x[0.40]^y   (4)

Dividing 4 by 3:\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}

2=2^x,2=2^1, x=1

Thus rate law is Rate=k[A]^1[B]^2

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1:  1.2\times 10^{-2}=k[0.10]^1[0.20]^2

k=3 L^2mol^{-2}s^{-1}.



6 0
3 years ago
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