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2 years ago

Please help asap!!!!!!!!!!!

Chemistry

1 answer:

Natali [406]2 years ago

4 0

Answer:

Explanation:

Where velocity doesn't change, THERE IS NO ACCEL....this is section B

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Write a Claim based in this question: What happens in the atmosphere and how

gregori [183]

Answer:

The atmosphere is the superhighway in the sky that moves water everywhere over the Earth. Water at the Earth's surface evaporates into water vapor which rises up into the sky to become part of a cloud which will float off with the winds, eventually releasing water back to Earth as precipitation.

Explanation:

If you don't want to plagiarize change it up a bit.

5 0

2 years ago

1. A student in lab measures 4.6 grams of copper for an experiment. Upon further analysis he determines that he should have meas

Reil [10]

Answer:

$2.13\%$

Explanation:

Quantity of copper measured by a student = 4.6 grams

Original quantity of copper = 4.7 grams

Error in measurement = Original quantity of copper - Quantity of copper measured by a student $=4.7-4.6=0.1$ grams

To find the percent error, apply the following formula:

Percent error = (Error in measurement / Original quantity of copper) × 100

$=\frac{0.1}{4.7}(100)=2.13\%$

4 0

4 years ago

What causes a gas to act in a non ideal manner?

il63 [147K]

"The forces of attraction and the volume of the molecules" (as opposed to the volume of the container the gas is in).

8 0

3 years ago

The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M. Based on thi

Grace [21]

Answer:

0.387 g

Explanation:

pH of the buffer = 1

V = Volume of solution = 100 mL

[HA] = Molarity of HA = 0.1 M

$K_a$ = Acid dissociation constant = $1.2\times 10^{-2}$

(assuming base as $Na_2SO_410H_2O$ )

Molar mass of base = 322.2 g/mol

pKa is given by

$pK_a=-\log K_a\\\Rightarrow pKa=-\log(1.2\times 10^{-2})\\\Rightarrow pK_a=1.92$

From the Henderson-Hasselbalch equation we get

$pH=pK_a+\log\dfrac{[A^-]}{[HA]}\\\Rightarrow pH-pK_a=\log\dfrac{[A^-]}{[HA]}\\\Rightarrow 10^{pH-pK_a}=\dfrac{[A^-]}{[HA]}\\\Rightarrow [A^-]=10^{pH-pK_a}[HA]\\\Rightarrow [A^-]=10^{1-1.92}\times0.1\\\Rightarrow [A^-]=0.01202\ \text{M}$