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3 years ago

A car travels at a constant speed of 65 mph. How far will it have traveled in 4 hours

Chemistry

2 answers:

Nostrana [21]3 years ago

6 0

It will have travelled 256 miles.

sattari [20]3 years ago

3 0

Ngl idk but just be a nice person and mark as brainliest please

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How many atoms are in a single molecule of FeCl2

Nady [450]

3
One from fe
And two from cl2

7 0

3 years ago

For the problem: Mg (s) + O₂ (g) ➞ MgO (g) if 4.3 g of magnesium reacts with 83 g of oxygen (O₂), which is the limiting reagent?

Anit [1.1K]

Explanation:

hope this can help youuu

3 0

3 years ago

What other signs provide evidence of chemical reactions and show that a chemical change has occurred?

Dmitry_Shevchenko [17]

Color change
new substance formed
heat or light change
new smell
irreversible

4 0

3 years ago

Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.

Rudiy27

For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3

Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol

m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg

For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf

Tf = -1.59 celsius

5 0

4 years ago

Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe

Illusion [34]

Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

$P_{B}=0.9999 atm$

Explanation:

For the reaction

A(g)⇌2B(g)

Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$