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lesantik [10]
3 years ago
11

Which of the following units is best when measuring the length of an object?

Chemistry
1 answer:
Marta_Voda [28]3 years ago
3 0

what kind of object?

if its super small ex. eraser.... use centimeter

if its pretty normal size like a peice of paper...use inches

if its a bit bigger than a foot or a foot...use feet/ foot

if its way bigger than a foot..use meter...

if its bigger than a meter..use yard

HOPE I HELPED

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Name three machines which convert straight motion to circular motion.
fgiga [73]
A Grinder,A Mixer,A Car 
3 0
3 years ago
how much heat will be absorbed by 25g of chloroform when its temperature started at 67.0C and after it was 112.0C
snow_tiger [21]

Hello friend ☺

ΔH = MCΔT

ΔH = to the amount of energy or change in energy (J)

mass of water

C = waters specific heat capacity

ΔT = change in temperature

and so ΔH = 25 × 4.18 × ( 112-67 ) J = 4702.5 J

Thanks ❤

8 0
3 years ago
What does atomic mass on the periodic table represent?
neonofarm [45]
The atomic mass on the periodic table represents the sum of number of protons and number of neutrons.

Atomic mass = Number of protons + number of neutrons

Hope this helps!
6 0
3 years ago
A sample of a hydrocarbon is found to contain 7.48g carbon and 2.52g hydrogen. What is the empirical
Aliun [14]

Answer:

CH₄

Explanation:

To determine the empirical formula of the hydrocarbon, we need to follow a series of steps.

Step 1: Determine the mass of the compound

The mass of the compound is equal to the sum of the masses of the elements that form it.

m(CxHy) = mC + mH = 7.48 g + 2.52 g = 10.00 g

Step 2: Calculate the percent by mass of each element

%C = mC / mCxHy × 100% = 7.48 g / 10.00 g × 100% = 74.8%

%H = mH / mCxHy × 100% = 2.52 g / 10.00 g × 100% = 25.2%

Step 3: Divide each percentage by the atomic mass of the element

C: 74.8/12.01 = 6.23

H: 25.2/1.01 = 24.95

Step 4: Divide both numbers by the smallest one, i.e. 6.23

C: 6.23/6.23 = 1

H: 24.95/6.23 ≈ 4

The empirical formula of the hydrocarbon is CH₄.

6 0
3 years ago
If iodine-131 has a half-life of 8 days, how much of a 64.0 g sample of iodine-131 will remain after 32 day?
Mariulka [41]

Answer:

4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.

Explanation:

Half life (t1/2) = 8 days

Original mass (No) = 64 g

Elapsed time (t) = 32 days

Mass remaining (Nt) = ?

Using the half life equation we can obtain the mass remaining (Nt)

Nt = No (1/2) ^t/t1/2

Substituting the values, we have;

Nt = 64 * ( 1/2 ) ^32/8

Nt = 64 * (1/2) ^4

Nt = 64 * 0.0625

Nt = 4 g

So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.

8 0
3 years ago
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