__Mg + __Fe2O3 -> __MgO + _

How much does the air weigh within a column that is 1 square inch in area that extends from sea level all the way to the top of

fredd [130]

Answer:

14.7 lbs

Explanation:

Air pressure is the weight of the air above us. It is approximately 14.7 pounds or lbs per square inch at sea level. It means that an air column weights 14.7 lbs, 1 square inch in diameter, reaching all the way up to the top of the atmosphere.

5 0

3 years ago

What point A on the phase diagram called?

Mariana [72]

Answer:

D) the critical point

Explanation:

Point A is the critical point in phase diagram. This is the highest temperature and pressure at which a pure material can exist in vapor/liquid equilibrium. Pretty cool!

6 0

3 years ago

rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is

Sunny_sXe [5.5K]

Complete Question

The rate of a certain reaction is given by the following rate law:

$rate = k [H_2][I_2]$

rate Use this information to answer the questions below.

What is the reaction order in $H_2$ ?

What is the reaction order in $I_2$ ?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in $H_2$ is n = 1

The reaction order in $I_2$ is m = 1

The overall reaction order z = 2

When the hydrogen is double the the initial rate is $rate_n = 4.0*10^{-4} M/s$

The rate constant is $k = 35.23 \ M^{-1} s^{-1}$

Explanation:

From the question we are told that

The rate law is $rate = k [H_2][I_2]$

The rate of reaction is $rate = 2.0 *10^{4} M /s$

Let the reaction order for $H_2$ be n and for $I_2$ be m

From the given rate law the concentration of $H_2$ is raised to the power of 1 and this is same with $I_2$ so their reaction order is n=m=1

The overall reaction order is

$z = n +m$

$z =1 +1$

$z =2$

At $rate = 2.0 *10^{4} M /s$

$2.0*10^{4} = k [H_2] [I_2] ---(1)$

= > $k = \frac{2.0*10^{4}}{[H_2] [I_2] }$

given that the concentration of hydrogen is doubled we have that

$rate = k [2H_2] [I_2] ----(2)$

=> $k = \frac{rate_n }{ [2H_2] [I_2]}$

So equating the two k

$\frac{2.0*10^{4}}{[H_2] [I_2] } = \frac{rate_n }{ [2H_2] [I_2]}$

=> $rate_n = 4.0*10^{-4} M/s$

So when

$rate_x = 52.0 M/s$

$[H_2] = 1.8 M$

$[I_2] = 0.82 \ M$

We have

$52 .0 = k(1.8)* (0.82)$

$k = \frac{52 .0}{(1.8)* (0.82)}$

$k = 35.23 M^{-2} s^{-1}$

3 0

3 years ago

Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- um

Andre45 [30]

Answer:

$\large \boxed{\text{0.0038 mol/L}}$

Explanation:

1. Calculate the initial moles of acid and base

$\text{moles of acid} = \text{0.0250 L} \times \dfrac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} \times \dfrac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}$

2. Calculate the moles remaining after the reaction

OH⁻ + H₃O⁺ ⟶ 2H₂O

I/mol: 0.0053 0.005 00

C/mol: -0.00500 -0.005 00

E/mol: 0.0003 0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

$\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}$

8 0

3 years ago

A test light with an electrical resistance of 9.00 S2 draws a current of 1.50 A when

yuradex [85]

Answer:

20.25 W

Explanation:

Applying,

P = I²R.................... Equation 1

Where P = Power, I = current drawn by the test light, R = Resistance of the test light

From the question,

Given: I = 1.5 A, R = 9.00 Ω

Substitute these values into equation 1

P = (1.5²)(9)

P = 2.25×9

P = 20.25 W

4 0

3 years ago

__Mg + __Fe2O3 -> __MgO + __Fe

Mg + Fe2O3 -> MgO + Fe