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hammer [34]
2 years ago
11

A student has 100. mL of 0.400 M CuSO4 (aq) and is asked to make 100. mL of 0.150 M CuSO4 (aq) for a spectrophotometry experimen

t. The following laboratory equipment is available for preparing the solution: centigram balance, weighing paper, funnel, 10 mL beaker, 150 mL beaker, 50 mL graduated cylinder, 100 mL volumetric flask, 50 mL buret, and distilled water.
Calculate the volume of 0.400 M CuSO4(aq) required for the preparation
Chemistry
2 answers:
GenaCL600 [577]2 years ago
5 0

Answer:

We have to take 37.5 mL of a 0.400 M solution

Explanation:

Step 1: Data given

Stock volume = 100 mL  = 0.100L

Stock concentration 0.400 M

Volume of solution he wants to make = 100 mL = 0.100L

Concentration of solution he wants to make = 0.150 M

Step 2: Calculate the volume of 0.400 M CuSO4 needed

C1*V1 = C2*V2

⇒with C1 = the stock concentration = 0.400M

⇒with V1 = the volume of the stock = TO BE DETERMINED

⇒with C2 = the concentration of the solution he wants to make = 0.150 M

⇒with V2 = the volume of the solution made = 0.100 L

0.400 M * V1 = 0.150M * 0.100L

V1 = (0.150M*0.100L) / 0.400 M

V1 = 0.0375 L = 37.5 mL

We have to take 37.5 mL of a 0.400 M solution

FromTheMoon [43]2 years ago
4 0

Answer:

37.5 mL

Explanation:

Step 1:

Data obtained from the question.

Concentration of the stock solution (C1) = 0.4M

Volume of stock solution needed (V1) =?

Concentration of diluted solution (C2) = 0.150 M

Volume of diluted solution (V2) = 100 mL

Step 2:

Determination of the volume of the stock solution needed to produce 100 mL of 0.150 M CuSO4.

This is illustrated below:

With the application of the diluted equation, the volume of the stock solution needed can be obtained as follow:

C1V1 = C2V2

0.4 x V1 = 0.150 x 100

Divide both side by 0.4

V1 = (0.150 x 100)/0.4

V1 = 37.5 mL

Therefore, 37.5 mL of the stock solution i.e 0.400 M CuSO4(aq) required for the preparation.

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Convert grams —> mols and then mols —> atoms

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Now we use avogadro’s number to do

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7 0
2 years ago
The volume of a single strontium atom is 4.15×10-23 cm3. What is the volume of a strontium atom in microliters
Ivenika [448]

Answer:-  4.15*10^-^2^0\mu L

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We know that:

1cm^3 = 1 mL

1mL=10^-^3L

and, 1L=10^6\mu L

Let's use these conversions factors for the desired conversion using dimensional as:

4.15*10^-^2^3cm^3(\frac{1mL}{1cm^3})(\frac{10^-^3L}{1mL})(\frac{10^6\mu L}{1L})

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7 0
2 years ago
The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s
denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0
2 years ago
A collection of coins contains 14 pennies, 16 dimes, and 7 quarters. What is the percentage of pennies in the collection?A colle
marin [14]

Answer:

=37.83783784

Explanation:

Find the total sum of all coins,

which is 37, take the number of pennies and the total of all coins put in parenthesis( 14/37) like so and than * times them by 100

you equation should look like this

(14/37)* 100= and than the answer shown above should be the one you received. I have checked this with multiple calculators, it should be accurate.  

8 0
2 years ago
Read 2 more answers
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vivado [14]

Answer:

Explanation:

wavelength λ = 12.4 x 10⁻² m .

energy of one photon = h c / λ

= 6.6 x 10⁻³⁴ x 3 x 10⁸ /  12.4 x 10⁻²

= 1.6 x 10⁻²⁴ J .

Let density of coffee be equal to density of water .

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heat required to heat up coffee = mass x specific heat x rise in temp

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No of photons required = heat energy required / energy of one photon

= 39438.3 / 1.6 x 10⁻²⁴

= 24649 x 10²⁴

= 24.65 x 10²⁷ .

5 0
2 years ago
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