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r-ruslan [8.4K]
3 years ago
7

Suppose you need

Chemistry
2 answers:
Reil [10]3 years ago
6 0
You calculate the mass using dimensional analysis.
We are given with
The weight of the chain per unit length which is 2.16 kg/m
We are also given with the needed length of the chain which is 7.0 m

Therefore, the mass of the chain is
7.0 m x 2.16 kg / m = mass of the chain in kg<span />
crimeas [40]3 years ago
4 0

Answer:

Mass of chain = \frac{216 kg}{1 m}  * \frac{7.0 m}{1}

Explanation:

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Uncooked lean ground beef can contain up to 12% fat by mass. How many grams of fat would be contained in 0.97 lbs of ground beef
klasskru [66]

52.8  grams of fat would be contained in 0.97 lbs of ground beef

8 0
3 years ago
Type the temperatures in Kelvin) in the left column
docker41 [41]

Answer: the data appear to be ( linear )

The line of best fit is v= 0.003 t+ 0.003

Notice the the value of the v intercept is close to 0 so V and t are ( directly proportional)

8 0
4 years ago
Read 2 more answers
Help for this practice work
allochka39001 [22]

Answer:

It's the third option.

Explanation:

In order for the chemical equation to be correctly it needs the same number of atoms of each element on both sides of the equal sign

8 0
2 years ago
Is boiling eggs chemical or physical practice?
Elden [556K]

Answer:

chemical

Explanation:

because heat is being taken to the egg

3 0
3 years ago
Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U
Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of nitrogen gas = 79mL/s

R_2 = rate of effusion of sulfur dioxide gas = ?

M_1 = molar mass of nitrogen gas  = 28 g/mole

M_2 = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}

R_2=52mL/s

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

4 0
3 years ago
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