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maksim [4K]
3 years ago
12

If you shake up different kinds or brands of soft drinks will they all spew you the same amount

Chemistry
2 answers:
Morgarella [4.7K]3 years ago
3 0

Answer:

yeah and the point?

Explanation:

Harman [31]3 years ago
3 0

Answer:

The different brands of sodas will not spew the same amount when shaken because there are different amounts of sugar in the sodas and that will affect the amount spewing. Pepsi will spew the highest, compared to the other brands of sodas

Explanation:

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What is the molarity of a solution containing 60. grams of NaOH dissolved in 400. mL of water?
Hunter-Best [27]

Answer: 3.75 M

Explanation:

400 mL = 0.4 L

NaOH has a molar mass of around 40 g/mol.

\frac{60 grams}{40 g/mol} = 1.5 moles

Molarity = \frac{1.5 moles}{0.4 Liters} = 3.75 M

8 0
2 years ago
The melting points and boiling points of 4 substances are shown.
Serggg [28]

Answer:

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C. with melting point 11° C and boiling point 181° C

Explanation:

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4 0
2 years ago
One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af
Licemer1 [7]

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0
3 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.
scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg


Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0
3 years ago
...................................................................Help dying of boredem
zhannawk [14.2K]
Can we talk here?
I don’t have what you want to talk on
5 0
3 years ago
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