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3 years ago

A precipitate will form when a freshly prepared aqueous carbonic acid solution is added to an aqueous solution of

Chemistry

1 answer:

galben [10]3 years ago

3 0

Answer:

A precipitate will form when a freshly prepared aqueous carbonic acid solution is added to an aqueous solution of Calcium Hydroxide.

Explanation:

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Two students mixed up a new salt solution and put in blue and red food coloring to make it purple. They weighed 35 mL of the sol

Gnoma [55]

Answer:

1435

Explanation:

i did the math and this is what i got i think its right but i dont know

7 0

3 years ago

Identify the atoms that are oxidized and reduced, the change in the oxidation state for each, and oxidising and reducing agents

KATRIN_1 [288]

Answer:

Explanation:

a) $Mg + NiCl_2 \rightarrow MgCl_2 + Ni$

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and $NiCl_2$ acts as oxidizing agent.

b) $PCl_3 + Cl_2 \rightarrow PCl_5$

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

$PCl_3$ acts as reducing agent and $Cl_2$ acts as oxidizing agent.

c) $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

$C_2H_4$ acts as reducing agent and $O_2$ acts as oxidizing agent.

d) $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and $H_2SO_4$ acts as oxidizing agent

e) $K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI$

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

$K_2S_2O_3$ acts as reducing agent and $I_2$ acts as oxidizing agent.

f) $3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO$

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and $HNO_3$ acts as oxidizing agent

8 0

2 years ago

Which of the following is true about matter

SashulF [63]

Answer: b- it is always a substance

Explanation: matter can be any liquid, gas, or solid that takes up space so the answer cannot be C or D. For answer number A to be true, you would have to remove the word more, because matter just has mass and takes up space it does not have "more mass"

8 0

3 years ago

Name the functional group in the<br> following molecule:

Anna71 [15]

Answer:

B. Aromatic

Explanation:

Functional groups are groups that differentiate a specific organic compound from others. A functional group determines the chemical property of the compound that possesses it.

For example, just like alkene and alcohol functional groups have characteristics double bond (=) and hydroxyl (OH) group respectively, the image in the attachment of this question has a BENZENE RING at the core of its structure, hence, the organic compound can be regarded to have an AROMATIC FUNCTIONAL GROUP.

4 0

3 years ago

Which is the correct net ionic equation for the following reaction: H3PO4 (aq) + 3 LiOH (aq) > Li3PO4 (aq) + 3 H2O (l)?

torisob [31]

Answer: The correct option for the chemical equation $H_3PO_4(aq.)+3LiOH(aq.)\rightarrow Li_3PO_4(aq.)+3H_2O(l)$ is 1.

The correct option for the chemical equation $2HBr(aq.)+Co(OH)_2(aq.)\rightarrow CoBr_2(aq.)+2H_2O(l)$ is 3.

Explanation:

For the chemical equation:

$H_3PO_4(aq.)+3LiOH(aq.)\rightarrow Li_3PO_4(aq.)+3H_2O(l)$

$H_3PO_4$ is a weak acid and hence will not dissociate into ions whereas LiOH is a strong base and will easily dissociate into ions.

The product $Li_3PO_4$ is soluble in water and hence, will dissociate into its respective ions. Hence, the ionic equation for this reaction is:

$H_3PO_4(aq.)+3Li^{3+}(aq.)+3OH^-(aq.)\rightarrow 3Li^{3+}(aq.)+PO_4^{3-}(aq.)+3H_2O(l)$

Net ionic equation becomes:

$H_3PO_4(aq.)+3OH^-(aq.)\rightarrow PO_4^{3-}(aq.)+3H_2O(l)$

So, the correct option is 1.

For the chemical equation:

$2HBr(aq.)+Co(OH)_2(aq.)\rightarrow CoBr_2(aq.)+2H_2O(l)$

HBr and $Co(OH)_2$ are strong acid and strong base respectively, hence they will easily dissociate into ions.

The product, $CoBr_2$ is is soluble in water and hence, will dissociate into its respective ions. Hence, the ionic equation for this reaction is:

$2H^+(aq.)+2Br^-(aq.)+Co^{2+}(aq.)+2OH^-(aq.)\rightarrow Co^{2+}(aq.)+2Br^-(aq.)+H_2O(l)$

The net ionic equation becomes:

$H^+(aq.)+OH^-(aq.)\rightarrow H_2O(l)$

So, the correct option is 3.

4 0

2 years ago

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