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artcher [175]
3 years ago
15

How many grams of magnesium metal will react completely with 6.3 liters of 4.5 M HCl? Show all of the work needed to solve this

problem.
Mg (s) + 2 HCl (aq) yields MgCl2 (aq) + H2 (g)
Chemistry
1 answer:
Rufina [12.5K]3 years ago
5 0
Mg (s) + 2 HCl (aq) -> MgCl2 (aq) + H2 (g)

1) Number of mols of HCl, n

n = 6.3 L * 4.5 mol/L = 28.35 mol

2) ratio: 2 mol HCl / 1 mol Mg = 28.35 mol HCl / x mol Mg

x = 28.35 mol HCL * 1 mol Mg / 2 mol HCL = 14.175 mol Mg

3) Convert mol to mass using atomic mass of Mg

14.175 mol Mg * 24.3 g Mg / mol Mg = 344. 45 g

Answer: 344.45 g

 

 


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A 27.3 g marble sliding to the right at 21.0 cm/s overtakes and collides with a 11.7 g marble moving in the same direction at 12
NISA [10]

Answer:

The answer to this is

The velocity of the 27.3Kg marble after collision is = 16.24 cm/s

Explanation:

To solve the question, let us list out the given variables and their values

Mass of first marble m1 = 27.3g

Velocity of the first marble v1 = 21.0 cm/s

Mass of second marble m2 = 11.7g

Velocity of the second marble v2 = 12.6 cm/s

After collision va1 = unknown and va2 = 23.7 cm/s

From Newton's second law of motion, force = rate of change of momentum produced

Hence m1v1 + m2v2 = m1va1 + m2va2 or

va1 = (m1v1 + m2v2 - m2va2)÷m2 or (720. 72-277.29)÷m1 → va1 = 16.24 cm/s

The velocity of the 27.3Kg marble after collision is = 16.24 cm/s

4 0
3 years ago
Plzzzzzzzzzzz help what is a font!
spayn [35]

Answer:

1. a receptacle in a church for the water used in baptism, typically a freestanding stone structure.

2. A type of writing or text style

Explanation:

There are mutiple definitions of font

6 0
3 years ago
A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°C. What is the magnitude of k at 95.0°C if Ea =
never [62]

Answer:

k ≈ 9,56x10³ s⁻¹

Explanation:

It is possible to solve this question using Arrhenius formula:

ln\frac{k2}{k1} = \frac{-Ea}{R} (\frac{1}{T2} -\frac{1}{T1} )

Where:

k1: 1,35x10² s⁻¹

T1: 25,0°C + 273,15 = 298,15K

Ea = 55,5 kJ/mol

R = 8,314472x10⁻³ kJ/molK

k2 : ???

T2: 95,0°C+ 273,15K = 368,15K

Solving:

ln\frac{k2}{k1} = 4,257

\frac{k2}{k1} = 70,593

{k2} = 9,53x10^3 s^{-1}

<em>k ≈ 9,56x10³ s⁻¹</em>

I hope it helps!

5 0
2 years ago
Air trapped in a cylinder fitted with a piston occupies 145 mL at 1.08 atm
pav-90 [236]

Answer: 0.0014 atm

Explanation:

Given that,

Original pressure of air (P1) = 1.08 atm

Original volume of air (T1) = 145mL

[Convert 145mL to liters

If 1000mL = 1l

145mL = 145/1000 = 0.145L]

New volume of air (V2) = 111L

New pressure of air (P2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1 = P2V2

1.08 atm x 0.145L = P2 x 111L

0.1566 atm•L = 111L•P2

Divide both sides by 111L

0.1566 atm•L/111L = 111L•P2/111L

0.0014 atm = P2

Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm

7 0
3 years ago
A. what is the hybridization of the central atom in cse2?
elixir [45]

The orbital hybridization of the central carbon atom in CSe2 is sp.

In chemical bonding, atomic orbitals may be combined to form appropriate hybrid orbitals suitable for bonding. The orbitals that combine during hybridization must be close enough in energy.

In the compound Cse2, carbon is the central atom bonded to two selenium atoms. The carbon atom in CSe2 is sp hybridized.

Learn more about orbital hybridization:  brainly.com/question/1869903

4 0
1 year ago
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