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3 years ago

What volume of a 1.3 M solution of Na2CO3 can be made from 800 ml. of a 4.9

Chemistry

1 answer:

galben [10]3 years ago

4 0

2.5 m that’s the answer yessir

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A 2.00 L sample of gas at 35C is to be heated at constant pressure until it reaches a volume of 5.25 L. To what Kelvin temperatu

Anastaziya [24]

Answer:

The sample will be heated to 808.5 Kelvin

Explanation:

Step 1: Data given

Volume before heating = 2.00L

Temperature before heating = 35.0°C = 308 K

Volume after heating = 5.25 L

Pressure is constant

Step 2: Calculate temperature

V1 / T1 = V2 /T2

⇒ V1 = the initial volume = 2.00 L

⇒ T1 = the initial temperature = 308 K

⇒ V2 = the final volume = 5.25 L

⇒ T2 = The final temperature = TO BE DETERMINED

2.00L / 308.0 = 5.25L / T2

T2 = 5.25/(2.00/308.0)

T2 = 808.5 K

The sample will be heated to 808.5 Kelvin

7 0

4 years ago

What is the molarity of a solution that contains 4.3 mol of acetic acid (C2H3O2) in 450. mL of solution?

lord [1]

Hey there !

Number of moles of solution: 4.3 moles

Volume in liters:

450.0 mL / 1000 => 0.45 L

Therefore:

Molarity = number of moles / volume of solution ( L)

Molarity = 4.3 / 0.45

=> 9.55 M

Hope that helps!

3 0

3 years ago

When Z-4,5-dimethyloct-4-ene is treated with hydrogen chloride, HCl, the result is:________.

Vika [28.1K]

Answer:

The correct option is c

Explanation:

The chemical equation for the reaction of Z-4,5-dimethyloct-4-ene and HCl is shown on the first uploaded image

Now looking at the product we see that there are two who has four different groups attached to them this carbon are known as chiral carbons hence the product formed is a pair of diastereomers

6 0

4 years ago

Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.

madreJ [45]

Answer:

$Molar \ solubility=3.12x10^{-5}M$

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

$CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-$

The equilibrium expression is:

$Ksp=[Ca^{2+}][F^-]^2$

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent $x$ is computed as follows:

$3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M$

Thus, the molar solubility equals the reaction extent $x$ , therefore:

$Molar \ solubility=3.12x10^{-5}M$

Regards.

4 0

3 years ago

What is a real life example of Charles’ law?

max2010maxim [7]

Leaving a basketball out in the cold weather. When a basketball if left in a cold garage or outside during the cold months, it loses its air inside (or volume).

6 0

3 years ago