4 years ago

< Practice Questions for Midterm 2 - Part 1 (Not Timed)

Chemistry

1 answer:

Sav [38]4 years ago

3 0

Answer: not helpin someone cheat

Explanation:

is this a test homie cuz if it is i cant help sorry

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Particle Mass (g) Atomic mass Electrical charge (C) Relative

KATRIN_1 [288]

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3 years ago

What is the volume of 3.50 moles of oxygen gas (O2) at 273 K and 1.00 atm?

faltersainse [42]

P x V = n x R x T

1.00 x V = 3.50 x 0.082 x 273

1.00 x V = 78.351

V = 78.351 / 1.00

V = 78.351 L

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3 years ago

A sample of a compound consisting only of copper and sulfur contains 88.39g of the metal and 44.61g of non-metal. How many grams

Lorico [155]

(88.39 / (88.39 + 44.61) ) x (5264000) = answer for copper grams

(44.61 / (88.39 + 44.61)) x (5264000) = answer for sulfur grams

Sulfur is a non-metal used to make paper, number 16 on periodic table.

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4 years ago

enny places a strip of pH paper into a solution. When she removes the pH paper, it has turned yellow-green. What should Jenny do

Svetllana [295]

Compare it to the chart showing how base or acidic a substance is

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4 years ago

In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t

Arada [10]

Answer:

$\large\boxed{\large\boxed{0.529M}}$

Explanation:

Since the rate constant has units of s⁻¹, you can tell that the order of the reaction is 1.

Hence, the rate law is:

$r=d[A]/dt=-k[A]$

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

$[A]=[A]_0e^{-kt}$

You know [A]₀, k, and t, thus you can calculate [A].

$[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}$

$[A]=0.529M$

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4 years ago