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3 years ago

How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 101.5 mL of 0.17 M H2SO4?

Chemistry

1 answer:

elena-14-01-66 [18.8K]3 years ago

6 0

Answer:

3.451 mL.

Explanation:

It is a dilution process.

We have the rule states that the no. of millimoles before dilution is equal to the no. of millimoles after dilution.

<em>(MV) before dilution = (MV) after dilution</em>

M before dilution = 5.0 M.

V before dilution = ??? mL.

M after dilution = 0.17 M.

V after dilution = 101.5 mL.

∴ The volume before dilution = (MV) after dilution / M before dilution = (0.17 M)(101.5 mL) / (5.0 M) = 3.451 mL.

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1.2 kg of aluminum at 20oC is added to 1.5 kg of water at 80oC. After the system reaches thermal equilibrium, what is its final

Lubov Fominskaja [6]

Answer:

The final temperature is 71.19 °C

Explanation:

Step 1: Data given

Mass of aluminium = 1.2 kg = 1200 grams

Temperature of aluminium = 20.0 °C

Specific heat of aluminium = 0.900 J/g°C

Mass of water = 1.5 kg = 1500 grams

Temperature of water = 80.0 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the final temperature

heat gained = heat lost

Q(aluminium) = - Q(water)

Q = m*c*ΔT

m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(water)

⇒ with mass of aluminium = 1200 grams

⇒ with specific heat of aluminium = 0.900 J/g°C

⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 20.0 °C

⇒ with mass of water = 1500 grams

⇒ with specific heat of water = 4.184 J/g°C

⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 80.0°C

1200 * 0.900 * (T2-20.0°C) = -1500 * 4.184 * (T2 - 80.0°C)

1080 * (T2 - 20.0°C) = -6276 * (T2 - 80.0°C)

1080 T2 - 21600 = -6276T2 + 502080

7356T2 = 523680

T2 = 71.19 °C

The final temperature is 71.19 °C

8 0

3 years ago

Which aqueous solution has a freezing point closer to its predicted value, 0.01 m NaBr or 0.01 m MgCl₂? Explain

emmasim [6.3K]

NaBr will behave more ideally and have an actual freezing point closer to its predicted (theoretical) value.

<h3>What is freezing point?</h3>

The temperature at which a liquid turns into a solid under normal air pressure is known as the freezing point.

NaBr in water dissociates to $Na^{+}$ and $Br^{-}$ (two particles) while $MgCl_{2}$ dissociating to $Mg^{2+}$ two $Br^{-}$ (three particles). There is a difference between a theoretical boiling/freezing point and an experimentally determined one and the primary cause of this phenomenon is the formation of positive ions on top of negative ones and vice versa, creating an ionic environment with a net opposite charge.

In such a setting we can conclude univalent ions will behave more ideally (bivalent ions have a 'greater' charge and are therefore affecting the solution interactions more). Also, since NaBr dissociates to only two particles, we can conclude it will behave more ideally. Finally, based on these two reasons, we can conclude that out of the two given compounds, NaBr will behave more ideally and have an actual freezing point closer to its predicted (theoretical) value.

To know more about freezing point, visit: brainly.com/question/3121416

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4 0

1 year ago

The specific heat of water is 4.2 J/g C. If a one gram sample of water absorbs 50 Joules of energy, how much will its temperatur

n200080 [17]

Answer: 12 g

Explanation: