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just olya [345]
3 years ago
9

A vial is filled with room temperature water, and blue cold water has been placed at the bottom. A warm water bag sits next to t

he vial. What do you think will happen?
Chemistry
1 answer:
MrRa [10]3 years ago
5 0

Explanation:

In the context, a vial which is used in store medical samples is filled with water at room temperature. And the vial is kept on a cold water. Also a water bag containing warm water is kept near the vial.

The cold water kept at the bottom of the vial is having lower kinetic energy while warm water will have higher kinetic energy than the others. Since the water in the vial is at room temperature and it is in touch with the cold blue water, the water in the vial will loose or give its temperature to the cold blue water through conduction as well as convection process since temperature always flows from a hot body towards the cold body.

On the other hand, the warm water placed next tot he vial will give its temperature to the atmosphere.  

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Desiré was investigating a chemical reaction.When she heated it up, she found that sulfuric acid changed into water. She made th
pochemuha

Answer:

no it is not a complete model

3 0
3 years ago
A saturated solution of baso4 has a concentration of 0.5mol/l. a 55ml sample is taken by you. what is the mass of baso4 in the s
SIZIF [17.4K]

Answer:

6.4 g BaSO₄

Explanation:

You have been given the molarity and the volume of the solution. To find the mass of the solution, you need to (1) find the moles BaSO₄ (via the molarity ratio) and then (2) convert moles BaSO₄ to grams BaSO₄ (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given values.

Molarity (mol/L) = moles / volume (L)

(Step 1)

55 mL / 1,000 = 0.055 L

Molarity = moles / volume                             <----- Molarity ratio

0.5 (mol/L) = moles / 0.055 L                        <----- Insert values

0.0275 = moles                                             <----- Multiply both sides by 0.055

(Step 2)

Molar Mass (BaSO₄): 137.33 g/mol + 32.065 g/mol + 4(15.998 g/mol)

Molar Mass (BaSO₄): 233.387 g/mol

0.0275 moles BaSO₄          233.387 g
---------------------------------  x  -------------------  =  6.4 g BaSO₄
                                                1 mole

6 0
2 years ago
A solution contains 0.115 mol h2o and an unknown number of moles of nacl. the vapor pressure of the solution at 30°c is 25.7 tor
Valentin [98]
According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr 
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808 
when we assume the moles of solute NaCl = X 
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
 So X (the no.of moles of NaCl) = 0.027 m
8 0
2 years ago
State Boyle's, Charles's, and Gay-Lussac's laws using sentences, then equations
dimaraw [331]

Explanation:

1. Boyle's Law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}      (At constant temperature and number of moles)

P_1\times {V_1}=P_2\times V_2

2. Charles' Law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T    (At constant pressure and number of moles

\frac{V_1}{T_1}=\frac{V_2}{T_2}

3. Gay Lussac's Law states that tempertaure is directly proportional to the pressure of the gas at constant volume and number of moles of gas

P\propto T    (At constant volume and number of moles)

\frac{V_1}{n_1}=\frac{V_2}{n_2}

7 0
2 years ago
To what Kelvin temperature must a balloon
snow_tiger [21]

Answer:

T_2=261.46\ K

Explanation:

It is given that,

Original temperature, T_1=323^{\circ}C=596.15\ K

Original volume, V_1=2.85\ L

We need to find the temperature if the volume of the balloon to be shrink to 1.25 L.

According to Charles law, at constant pressure, V\propto T

It would means, \dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}

T₂ = ?

T_2=\dfrac{V_2T_1}{V_1}\\\\T_2=\dfrac{1.25\times 596.15}{2.85}\\\\T_2=261.46\ K

So, the new temperature is 261.46 K.

4 0
3 years ago
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