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dangina [55]
3 years ago
8

Dr. Martin is an ophiologist, or a scientist who studies snakes. During one experiment, Dr. Martin fed a snake a whole mouse and

compared the mass of the snake before it consumed the mouse to the snake's mass immediately after it was fed. According to the law of conservation of mass, how should the masses compare?
A. The mass of the snake after feeding should be the same as the original mass of the snake.
B. The mass of the snake after feeding should be equal to the mass of the mouse. C. The mass of the snake after feeding should be equal to the original mass of the snake minus the mass of the mouse.
D. The mass of the snake after feeding should be equal to the original mass of the snake plus the mass of the mouse.
Chemistry
2 answers:
sveticcg [70]3 years ago
6 0

Answer:

D

Explanation:

I just did it on study island

Anestetic [448]3 years ago
4 0

D) mass of snake after feeding is equal to orginal mass of snake plus mouse since the mass is conserved

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Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
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8 0
2 years ago
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BabaBlast [244]

Answer:

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Solution:

Data Given:

                             Mass  =  42 g

                             Volume  =  35 cm³

Formula Used:

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Putting values,

                             Density  =  42 g ÷ 35 cm³

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Answer:

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