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Klio2033 [76]
3 years ago
12

Convert the pressure 525.4 torr to kPa.

Chemistry
2 answers:
Kruka [31]3 years ago
6 0
The awnser is A - 70.0
Oduvanchick [21]3 years ago
3 0

Answer:

Option A. 70.0 KPa.

Explanation:

Data obtained from the question include:

Pressure (torr) = 525.4 torr

Pressure (kPa) =?

The pressure expressed in torr can be converted kPa as shown below:

760 torr = 101.325 KPa

Therefore,

525.4 torr = (525.4 x 101.325) / 760 = 70.0 KPa.

Therefore, 525.4 torr is equivalent to 70.0 KPa.

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4 0
3 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
3 years ago
1. Given the specific heat of lead is 0.129 J/g.C and that it takes 93.4J of energy to
zheka24 [161]

Answer: 40 grams

Explanation:

The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since Q = 93.4J

M = ?

C = 0.129 J/g.C

Φ = 40.4°C - 22.3°C = 18.1°C

Then, Q = MCΦ

Make Mass, M the subject formula

M = Q/CΦ

M = (93.4J) / (0.129 J/g.C x 18.1°C)

M = 93.4J / 2.33J/g

M = 40 g

Thus, the mass of the lead is 40 grams

8 0
3 years ago
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