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Rudiy27
3 years ago
12

A can of soda has a volume of 250 mL and a mass of 100 g? what is the density of the soda

Chemistry
1 answer:
Doss [256]3 years ago
3 0

0.4 g/ml.............................

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What does CoH12O6 tell you about the glucose molecule?
Leokris [45]

Answer:

Glucose = C6H12O6

molecular mass = 6(12) + 12(1) + 6(16)

= 72 + 12 + 96

= 180 g

Explanation:

Glucose has a chemical formula of: C6H12O6 That means glucose is made of 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms. ... Glucose is produced during photosynthesis and acts as the fuel for many organisms.

8 0
3 years ago
According to the bohr model of the atom, where can atoms exist? where will they ot exist
Inga [223]
At least, that's what Bohr<span> decided, and that's why he proposed the </span>existence<span> of the</span>atomic<span> energy level. </span>According<span> to </span>Bohr<span>, the electrons in an </span>atom<span> were only allowed to </span>exist<span> at certain energy levels</span>
6 0
3 years ago
Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio
umka21 [38]

Answer : The value of K_p at this temperature is 66.7

Explanation : Given,

Pressure of PCl_3 at equilibrium = 0.348 atm

Pressure of Cl_2 at equilibrium = 0.441 atm

Pressure of PCl_5 at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}

Now put all the values in this expression, we get :

K_p=\frac{(10.24)}{(0.348)(0.441)}

K_p=66.7

Therefore, the value of K_p at this temperature is 66.7

4 0
3 years ago
how many calories is in one peanut if the volume of water is 10 mL and the water temperature is rise 4 degrees celsius
liraira [26]

Answer:

40.02 calories

Explanation:

V = 10 mL = 10g

we know t went <em>up</em> by 4°C, this is our ∆t as it is a change.

Formula that ties it together: Q = mc∆t

where,

Q = energy absorbed by water

m = mass of water

c = specific heat of water (constant)

∆t = temperature change

Q = (10 g) x (4.186 J/g•°C) x (4°C)

Q = 167.44 J

Joules to Calories:

167.44 J x 1 cal/4.184 J = 40.02 calories

(makes sense as in image it is close to the value).

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