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Gnesinka [82]
3 years ago
5

I need answer for a. (ii) pls​

Chemistry
1 answer:
allsm [11]3 years ago
4 0

Answer:

Stopwatch

Stopwatch is a time measuring tool.

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<span>A metal is one which ionizes easily and gives electrons whencompared to other elements. So it should have a lower ionization energy. Hence we would expect element 2 to be a metal.</span>
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A student wrote The Following hypothesis if more salt is added to a hypothesis
Hatshy [7]

Answer:

i love adding salt to my hypothesis

Explanation:

4 0
3 years ago
Compare the charges and masses of protons, neutrons, and electrons.
Eva8 [605]

Answer:

Explanation:

Protons have a positive charge, with a mass of 1 amu.

Neutrons have no charge, also with a mass of 1 amu.

Electrons have a negative charge, with a mass of 1 amu.

This is why when you're calculating the mass of an element or isotope, we only count the number of protons and neutrons.

3 0
3 years ago
The mole fraction of NaCl in an aqueous solution is 0.132. How many moles of water are present in 1 mole of this solution?
SOVA2 [1]
You could look at it this way.
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answer: 0.868 <<<< answer


7 0
3 years ago
Read 2 more answers
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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