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3 years ago

How to know if you can heat a laboratory flask and other glassware?

Chemistry

1 answer:

AnnyKZ [126]3 years ago

7 0

Answer:

fill the flask with water. apply heat, when you see the vapor you will know is heat

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What is the degree of oxidation of a simple substance

a_sh-v [17]

The oxidation state of a free element (uncombined element) is zero. For a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion. For example, Cl– has an oxidation state of -1. When present in most compounds, hydrogen has an oxidation state of +1 and oxygen an oxidation state of −2.

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3 years ago

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Which contributes to the polarity of a water molecule?

Verizon [17]

It’s molecular shape.

When the two hydrogen atoms bond with the oxygen, they attach to the top of the molecule rather like Mickey Mouse ears. This molecular structure gives the water molecule polarity, or a lopsided electrical charge that attracts other atoms. The end of the molecule with the two hydrogen atoms is positively charged.

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3 years ago

The carbon atom in co2 shares _______blank electron pairs. after the reaction with h2o to form h2co3 is completed, the carbon at

lisov135 [29]

B , four , four ........

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3 years ago

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A mixture of NaCl and CuCO3 was dissolved in water and the resulting slurry was filtered in to a beaker. The beaker and the filt

Firdavs [7]

The percentage adds up to 100%.

<h3>Percentages</h3>

Mass of mixture = mass of container+mixture - mass of container = 56.779 - 54.558 = 2.221 g

CuCO3 is insoluble in water. Thus:

Mass of CuCO3 = mass of beaker and residue - mass of beaker = 78.875 - 77.575 = 1.300 g

Mass of NaCl = mass of mixture - mass of CuCO3 = 2.221 - 1.300 = 0.921 g

%NaCl (w/w) = weight of NaCl/weight of mixture = 10.921/2.221 = 41.468%

% (w/w) CuCO3 = weight of CuCO3/weight of mixture = 1.300/2.221 = 58.532%

Addition of percentages = 41.468 + 58.532 = 100%

More on percentages can be found here: brainly.com/question/24159063

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2 years ago

g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each

RUDIKE [14]

Answer:

For a: Lead iodide is a yellow precipitate.

For b: Barium sulfate is a white precipitate.

For c: Ferric hydroxide is a brown precipitate.

For d: Copper (II) hydroxide is a blue precipitate.

Explanation:

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given options:

For (a):

The chemical reaction between KI and lead (II) nitrate follows:

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)$

The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.

For b:

The chemical reaction between barium chloride and sulfuric acid follows:

$BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)$

The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.

For c:

The chemical reaction between NaOH and ferric chloride follows:

$3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)$

The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.

For d:

The chemical reaction between NaOH and copper sulfate follows:

$CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4$

The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.

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3 years ago