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3 years ago

3. Calculate the answers to the appropriate number of significant figures. e) 43.678 x 64.1 = f) 1.678/0.42 =

Chemistry

1 answer:

Sophie [7]3 years ago

8 0

Where’s the pictures

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(01.07 но) An experiment was conducted to determine the density of a liquid. The table shows a partial record of the experiment

Anestetic [448]

Answer:

How do you find the density of a liquid experiment?

To measure the density of a liquid you do the same thing you would for a solid. Mass the fluid, find its volume, and divide mass by volume. To mass the fluid, weigh it in a container, pour it out, weigh the empty container, and subtract the mass of the empty container from the full container.

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2 years ago

At what temperatures is a reaction that has a positive change in entropy

Setler [38]

Answer:

the answer is A

I made a chart for AP chem if you want to refer to it.

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3 years ago

Private land is generally used for _______. a. homes b. businesses c. factories d. all of the above Please select the best answe

alekssr [168]

HM, I think the answer would be D. This is just a guess, so please use it if ou want to answer D it's ok :D

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3 years ago

Read 2 more answers

HELP PLS PLS PLS ILL MARK U BRAINLIEST

777dan777 [17]

Answer:

2 FeCl3 → 2 Fe + 3 Cl2

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2 years ago

A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

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3 years ago