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Troyanec [42]
3 years ago
12

Starting with copper metal, describe how to prepare a pure sample of copper (II) carbonate.

Chemistry
1 answer:
adoni [48]3 years ago
7 0

Answer:

I want brainliest plz!!!

Explanation:

React excess copper metal with concentrated sulphuric acid or concentrated nitric acid until effervescence stops. Filter to remove excess copper. To the filtrate add either sodium carbonate or potassium carbonate to form copper (II) carbonate, to precipitate copper (II) carbonate wash and dry between filter paper

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How does the structure of water explain its high boiling point, high heat capacity, and high heat of vaporization?
nikklg [1K]
Water's high heat capacity<span> is a property caused by hydrogen bonding among </span>water<span> molecules. When </span>heat<span> is absorbed, hydrogen bonds are broken and </span>water <span>molecules </span>can<span> move freely. When the temperature of </span>water decreases, the hydrogen bonds are formed and release a considerable amount of energy.
<span>Water's heat of vaporization is around 540 cal/g at </span>100 °C<span>, water's boiling point.

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7 0
3 years ago
Match the prefixes. 1. 1 hex- 2. 2 eth- 3. 3 prop- 4. 4 hept- 5. 5 non- 6. 6 dec- 7. 7 pent- 8. 8 but- 9. 9 meth- 10. 10 oct-
erma4kov [3.2K]

Hi!

All these prefixes are commonly employed in the nomenclature in organic chemistry. These are associated with organic compounds, such as alkanes, alkenes, alcohols and carboxylic acids -and these prefixes are known as word root.

<h3>The answers would be:</h3><h3>a) 1 - meth</h3>

Meth is associated with the number one, in that it signifies the presence of one carbon in the compound in concern. For instance, a one carbon alkane would be known as methane -with <em>ane </em>being indicative of the compound being an alkane, and meth denoting the presence of one carbon

<h3>b) 2 - eth</h3>

Eth is the associated prefix with the number 2 as it denotes the presence of 2 carbon atoms in the compound. For instance, the compound ethene gets it name for having 2 carbon atoms (denoted by eth). The eth acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>c) 3 - prop</h3>

Prop is the prefix that signifies the presence of 3 carbons in the compound in concern. For instance, propanol is a three carbon alcohol, which we can tell by the name. Prop indicating three carbons in the compound, and <em>anol</em> being indicative of the compound being an alcohol.

<h3>d) 4 - but</h3>

But is associated with the number one, in that it denotes the presence of 4 carbons in the compound. For instance, a 4 carbon carboxylic acid would be known as butanoic acid -with anoic<em> </em>being indicative of the compound being a carboxylic acid, and but signifying the presence of four carbons.

<h3>e) 5 - pent</h3>

Pent is the associated prefix with the number five as it denotes the presence of five carbon atoms in the compound. For instance, the compound pentene gets it name for having five carbon atoms (denoted by pent). The pent acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>f) 6 - hex</h3>

Hex is the prefix that signifies the presence of six carbons in the compound in question. For instance, a six carbon alkane would be known as hexane -with <em>ane </em>being indicative of the compound being an alkane, and hex denoting the presence of one carbon

<h3>g) 7 - hept</h3>

Hept is the associated prefix with the number seven as it denotes the presence of seven carbon atoms in the compound. For instance, the compound heptene gets it name for having seven carbon atoms (denoted by hept). The hept acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>h) 8 - oct</h3>

Oct is the prefix that is associated with the number 8. This is because it denotes the presence of eight carbon atoms in the compound. For instance, the compound octane gets it name for having eight carbon atoms (denoted by oct). The oct acts as a prefix to the characteristic <em>ane</em> of alkanes.

<h3>i) 9 - non</h3>

Non is the prefix that is associated with the number nine, and thus denotes the presence of nine carbon atoms in the compound. For instance, for alkanes, a nonane would be a nine carbon atom alkane, with non denoting the presence of nine carbon atoms, and <em>ane</em> being indicative that the compound it an alkane.

<h3>j) 10- dec</h3>

Dec is the prefix that is associated with the number nine, and thus denotes the presence of nine carbon atoms in the compound. For instance, for alkanes, a decane would be an alkane with ten carbon atoms, with dec denoting the presence of ten carbon atoms, and <em>ane</em> being indicative that the compound it an alkane.

<h3>Hope this helps!</h3>
4 0
3 years ago
Which property of light does this diagram show?
lidiya [134]
Full question need, post more
6 0
3 years ago
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO
Irina-Kira [14]

Answer:

Amount of excess Carbon (ii) oxide left over = 23.75 g

Explanation:

Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

Molar mass of Fe₂O₃ = 160 g/mol;

Molar mass of Carbon (ii) oxide = 28 g/mol

From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g)  of carbon (ii) oxide

450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide

Therefore the excess reactant is carbon (ii) oxide.

Amount of excess Carbon (ii) oxide left over = 260 - 236.25

Amount of excess Carbon (ii) oxide left over = 23.75 g

5 0
3 years ago
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