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finlep [7]
2 years ago
8

One of the products when aqueous Na, CO, reacts with aqueous

Chemistry
1 answer:
mixas84 [53]2 years ago
6 0

Answer:

NaNO3

balanced equation - Na2CO3 + Sn(NO3)2 → 2NaNO3 + Sn(CO3)

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Which kind of covalent bond has unequal sharing of electrons
lutik1710 [3]

Polar.

Polar bonds have unequal sharing electrons while nonpolar, the opposite, has equal sharing electrons. This is a tactic typically used to determine whether or not a compound or element itself is polar or nonpolar.

Hope this helps!

7 0
3 years ago
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In acid solution, water can add to the double bond of 2-butenedioic acid to form 2-hydroxysuccinic acid.
satela [25.4K]
So, what is the question?
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3 years ago
The order of a reaction Z1) is the product of the powers to which the reactant concentrations are raised in the rate law. Z2) ca
irinina [24]

Explanation:

The Order of Reaction refers to the power dependence of the rate on the concentration of each reactant.

The overall order of reaction is the sum of the individual orders of reaction with respect to the reactants.

Rate = k [A]²[B]¹

In the rate law above, the rate is second order with respect to A and first order with respect to B. The overall order of reaction is a third order reaaction given as; 2+ 1 = 3

6 0
4 years ago
What do the numbers listed for an element on the periodic table represent?
Anna007 [38]
The smaller number is the number of protons, and the greater number is the mass.
8 0
3 years ago
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Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).
wolverine [178]

Answer: Li is the reducing agentg and O is the oxidizing agent.

Explanation:

1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.

2) The given reaction is:

4Li(s) + O₂ (g) → 2 Li₂O(s)

3) Determine the oxidation states of each atom:

Li(s): oxidation state = 0 (since it is alone)

O₂ (g): oxidation state = 0 (since it is alone)

Li in Li₂O (s) +1

O in Li₂O -2

That because 2× (+1) - 2 = 0.

4) Determine the changes:

Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.

O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.

7 0
4 years ago
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