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BigorU [14]
3 years ago
13

A(n) ______ is a mixture that does not scatter light.

Chemistry
2 answers:
Mice21 [21]3 years ago
7 0

Explanation:

colloid is the answer but iam tired to write explanation

34kurt3 years ago
7 0
Colloid is the answer
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Consider the above unbalanced equation. What volume of CO2 is produced at 270. mm Hg and 38.5°C when 0.820 g of C4H8 reacts with
irinina [24]

Answer:

The volume of CO2 is 4.20 L

Explanation:

Step 1: Data given

Pressure = 270 mm Hg =  260 /760 = 0.355263 atm

Temperature : 38.5 °C = 311.65 K

Mass of C4H8 = 0.820 grams

Step 2: The balanced equation

C4H8 + 6O2 → 4CO2 + 4H2O

Step 3: Calculate moles C4H8

Moles C4H8 = mass C4H8 / molar mass C4H8

Moles C4H8 = 0.820 grams / 56.11 g/mol

Moles C4H8 = 0.0146 moles

Step 4: Caalculate moles CO2

For 1 mol C4H8 we need 6 moles O2 to produce 4 moles CO2 and 4 moles H2O

For 0.0146 moles we'll have 4*0.0146 = 0.0584 moles CO2

Step 6: Calculate Volume CO2

p*V = n*R*T

V = (n*R*T) /p

⇒ with V = the volume of CO2 = TO BE DETERMINED

⇒ with n = the moles of CO2 = 0.0584 moles

⇒ with R = the gas constant = 0.08206 L*atm / mol*K

⇒ with T = The temperature = 311.65 K

⇒ with p = the pressure = 0.355263 atm

V = (0.0584 * 0.08206 * 311.65) / 0.355263

V = 4.20 L

The volume of CO2 is 4.20 L

8 0
3 years ago
PLEASE HELP WITH #1 AND #2 ASAP!! PLEASE
Grace [21]
1. Since the track is extended outwards from lane one, the other lanes will be a longer distance then lane one. So the runners in the lanes (2,3,4,5,6,7,8) will start ahead of the runner in lane one so they all run the same distance. 2. The displacement of runner 1 would be that since they are farther back then they will have to run top speed in order to catch up to the other runners, especially runner 8 since it seems like they are so far ahead. This isn’t necessarily the same for all the runners because runner 1 is the farthest back and is at a disadvantage since they are in the inner lane and not the outer lane. I reallyyyy hope this helped :)
5 0
3 years ago
1. Determine the number of significant figures in the following values:
denpristay [2]

Answer:

The zero to the left of a decimal value less than 1 is not significant.All trailing zeros that are placeholders are not significant.Zeros between non-zero numbers are significant.All non-zero numbers are significant.

7 0
2 years ago
Assume that a milliliter of water contains 20 drops. How long, in hours, will it take you to count the number of drops
garri49 [273]

Answer:

126.18 hr

Explanation:

Data given:

1 mL of water = 20 drops

count rate = 10 drops/s

time in hours for one gallon = ?

Solution:

First we calculate number of mL (milliliter) of water in gallon

As we know

1 gallon = 3785.4 mL

As,

1 galon consist of 3785.4 mL of water, so now we count number of drops that contain 3785.4 mL of water

As we Know 1 mL water contain 20 drops then 3785.4 mL of water contain how many drops:

Apply unity formula

                    1 mL water ≅ 20 drops

                    3785.4 mL water ≅ X drops

Do cross multiplication

                 X drops of water = 20 drops x 3785.4 mL / 1 mL

                 X drops of water = 75708 drops

So, we come to know that one gallon contain 75708 drops of water and we have to calculate the time in hour to count these drops

First we calculate time in seconds

As we Know 10 drops water count in one second then how many seconds it will take to count 75708 drops

Apply unity formula

                    1 second ≅ 10 drops

                    X second ≅ 75708 drops

Do cross multiplication

                 X second  = 1 second x 75708 drops / 10 drops

                 X second = 7570.8 second

So it take 7570.8 second to count 1 gallon water drops

Now convert seconds to hours

As,

60 seconds = 1 hr

7570.8 second  =  7570.8 / 60 = 126.18 hr

So it take 126.18 hr to count 1 gallon water drops.

6 0
4 years ago
During oxidation, which of the following is true? A. Electrons are gained, so the oxidation number increases. B. Electrons are g
EastWind [94]

Oxidation is when the overall charge (or oxidation number) increases.  The only way to increase an oxidation number is to lose an electron, thereby making the negative charges less.  The correct answer is C.

6 0
3 years ago
Read 2 more answers
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