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3 years ago

HELP PLEASE. 50 points!

Chemistry

2 answers:

Lunna [17]3 years ago

4 0

Answer: No e nada sobre ato pero gracias por los puntos

Explanation:

devlian [24]3 years ago

4 0

Answer:

The image is way to small i cant read it

Explanation:

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Some distilled water is added to an empty beaker. a gram of copper (ii) nitrate is added to the beaker and while the water is be

Xelga [282]

Copper ions, nitrate ions, and water is in the beaker

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3 years ago

Which of these functional groups is found in all amino acids? -NH2 -OH =O -PO42-

melisa1 [442]

Answer : The correct option is, $-NH_2$

Explanation :

Amino acid : The acid that contains two functional groups that are carboxylic group, $-COOH$ and ammine group, $-NH_2$ .

When the two or more that two amino acids join together with the help of peptide bond, they produces polypeptide chain or protein.

The bond present between the two amino acid is called a peptide bond.

The peptide bond is a chemical bond that is formed between the two molecules when the nitrogen of one amino acid react with the carbon of another amino acid by releasing a water molecule. This is a dehydration synthesis or condensation reaction.

From this we conclude that, only two functional groups carboxylic group, $-COOH$ and ammine group, $-NH_2$ are present in all amino acids.

Hence, the correct option is, $-NH_2$

7 0

3 years ago

How many moles of silver can be produced from silver nitrate from 1 mole of zinc?

jasenka [17]

Answer:

Answer: 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

Explanation:

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

Zn + 2AgNO3 ---> Zn (NO3)2 +2Ag

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = 1/2 x 6.5 = 3.25 moles of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal, hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = 2/2 x 6.5 = 6.5moles of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

4 0

2 years ago

A balloon containing 0.0400 mol of a gas with a volume of 500 mL was expanded to 1.00 L. Answer the questions and round answers

grandymaker [24]

the number of moles 0.08

8 0

3 years ago

Phosphorous pentoxide, P2O5(s), is produced from the reaction between pure oxygen and pure phosphorous (P, solid). What is the v

BartSMP [9]

Answer:

4190.22 L = 4.19 m³.

Explanation:

For the balanced reaction:

2P₂ + 5O₂ ⇄ 2P₂O₅.

It is clear that 2 mol of P₂ react with 5 mol of O₂ to produce 2 mol of P₂O₅.

Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:

no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.

Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:

Using cross multiplication:

5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.

Finally, we can get the volume of oxygen using the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 60.95 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.

3 0

3 years ago