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Doss [256]
3 years ago
5

What is mA to the nearest degree

Chemistry
1 answer:
IrinaVladis [17]3 years ago
8 0

the answer to this would be that AB if BC= 11m and mA is 54 degrees.

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Question 2(Multiple Choice Worth 3 points)
viktelen [127]

Answer:

2.0 x 10⁻³ M/s.

Explanation:

  • The rate of the reaction is the change of the concentration of the reactants or the products with time.

<em>The rate of the reaction = - Δ[reactants]/Δt = - [(0.6 M - 1.8 M)]/(580 s - 0 s) = 2.069 x 10⁻³ M/s.</em>

5 0
3 years ago
(AKS 2e) Group 2 metals bond with nonmetals or polyatomic ions. What question
Lyrx [107]

Answer:

Will group 2 elements lose electrons to bond with nonmetals in group 17 in a

1:2 ratio?

Explanation:

Metals are electropositive in nature. This means that they loose electrons. Thus, metals form ionic bonds by loosing electrons to non metals.

Elements of group 2 have a valency of 2 while those of group 17 has a valency of 1 so the ratio in which group 2 elements bond with elements of group 17 is 1:2. Hence the answer.

3 0
3 years ago
In physical changes, the atoms or molecules that compose the matter do not change their identity, even though the matter may cha
aalyn [17]
True. Physical changes only affect the inter-molecular bonding and structure.
4 0
3 years ago
Air trapped in a cylinder fitted with a piston occupies 145.7 mL at 1.08 atm pressure. What is the new volume when the piston is
Anon25 [30]

The new volume of air trapper in the cylinder is determined as 116.56 mL.

<h3>New volume of air in the cylinder</h3>

The volume of air trapper in the cylinder is determined from Boyle's law;

P₁V₁ = P₂V₂

where;

  • P₁ is initial pressure = 1.08 atm
  • V₁ is initial volume = 145.7 mL
  • P₂ is new pressure = 1.25 x 1.08 atm = 1.35 atm
  • V₂ is new volume = ?

V₂ = (P₁V₁)/P₂

V₂ = (1.08 x 145.7)/1.35

V₂ = 116.56 mL

Thus, the new volume of air trapper in the cylinder is determined as 116.56 mL.

Learn more about volume here: brainly.com/question/1972490

#SPJ1

6 0
2 years ago
What is the answer for those questions plz ,help
marysya [2.9K]

Answer:

1) 1.51 × 10²³  particles of Mg

2) 0.54 × 10⁻³  moles

Explanation:

Given data:

1)

Number of moles of Mg = 0.250 mol

Number of representative particles = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mol = 6.022 × 10²³  particles

0.250 mol × 6.022 × 10²³  particles / 1 mol

1.51 × 10²³  particles of Mg

2)

Given data:

Number of moles of lead = ?

Number of atoms of lead = 3.25×10²⁰ atoms

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mol = 6.022 × 10²³  atoms

1 mol × 3.25×10²⁰ atoms/ 6.022 × 10²³  atoms

0.54 × 10⁻³  moles

6 0
3 years ago
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