All categories

3 years ago

To 25mL of 0.90M HCl, 275mL do distilled water is added. What is the molarity of the resulting solution?

Chemistry

1 answer:

nalin [4]3 years ago

8 0

Answer:

0.075 M.

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 = 0.90 M.

V before dilution = 25.0 mL.

M after dilution = ??? M.

V after dilution = V of HCl + V of water added = 25.0 mL + 275.0 mL = 300.0 mL.

∴ M after dilution = (MV) before dilution / V after dilution = (0.90 M)(25.0 mL) / (300.0 mL) = 0.075 M.

You might be interested in

A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.

Maslowich

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

5 0

3 years ago

Be sure to answer all parts. Select the anti periplanar geometry for the E2 reaction of (CH3)2CHCH2Br with base. mc2x mc2xd1 mc2

sleet_krkn [62]

The anti periplanar geometry for the E2 reaction of (CH3)2CHCH2Br with base is shown in the image attached as well as the structure of the product formed in the reaction.

In organic chemistry, an antiperiplanar conformation is one in which the groups point up and down at a dihedral angle of 180° away from one another. In the image attached, the antiperiplanar conformation of (CH3)2CHCH2Br is shown.

Recall that an E2 reaction is a synchronous elimination reaction where to atoms leave at the same time. The product of this reaction is also shown in the image attached.

Learn more: brainly.com/question/2510654