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Yuki888 [10]
2 years ago
10

How many grams (g) are in 0.250 moles of copper?

Chemistry
1 answer:
MissTica2 years ago
5 0

Answer:

15.89grams

Explanation:

The mass of a substance can be calculated from it's mole value by using the formula:

mole = mass/molar mass

According to this question, there are 0.250 moles of copper. Hence, the mass of copper can be calculated as follows:

Molar mass of Cu = 63.55g/mol

0.250 = mass/63.55

mass = 0.250 × 63.55

mass = 15.8875

Mass of Cu in 0.250mol is 15.89grams.

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Which two substances are covalent compounds? * C6H12O6(s) and KI(s) C6H12O6(s) and HCl(g) KI(s) and NaCl(s) NaCl(s) and HCl(g)
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Answer: Option (b) is the correct answer.

Explanation:

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For example, C_{6}H_{12}O_{6} and HCl is also a covalent compound.

And, a compound in which transfer of electrons occur between the combining atoms is known as an ionic compound. Whenever, a metal chemically combines with a non-metal then it will always lead to the formation of an ionic compound.

For example, KI is an ionic compound.

Thus, we can conclude that C_{6}H_{12}O_{6} and HCl are the two substances which are covalent compounds.

7 0
3 years ago
The decomposition of HBr(g) into elemental species is found to have a rate constant of 4.2 ×10−3atm s−1. If 2.00 atm of HBr are
Dennis_Churaev [7]

Answer:

7,94 minutes

Explanation:

If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>

For the zero-order reactions, concentration-time equation can be written as follows:

                                          [A] = - Kt + [Ao]

where:

  • [A]: concentration of the reactant A at the <em>t </em>time,
  • [A]o: initial concentration of the reactant A,
  • K: rate constant,
  • t: elapsed time of the reaction

<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>

Data:

K = 4.2 ×10−3atm/s,  

[A]o=[HBr]o= 2 atm,  

[A]=[HBr]=0 atm (all HBr(g) is gone)

<em>We clear the incognita :</em>

[A] = - Kt + [Ao]............. Kt =  [Ao] - [A]

                                        t  = ([Ao] - [A])/K

<em>We replace the numerical values:</em>

t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes

So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).

6 0
3 years ago
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