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SOVA2 [1]
2 years ago
15

A gem has a mass of 5. 50 g. When the gem is placed in a graduated cylinder containing 9. 500 ml of water, the water level rises

to 10. 45 ml
Chemistry
1 answer:
Katena32 [7]2 years ago
6 0

The density in grams per milliliter  7.975.

<h3>What is density?</h3>

Density is the mass of per unit volume. Density is equals to mass into volume.

D = M x V

The mass of a gem is given, M = 5.50

The volume is 10.45 - 9.00 = 1.45

Thus, the density is

D = 5.50 × 1.45 = 7.975

Thus, the density in grams per milliliter is 7.975.

Learn more about density

brainly.com/question/19320982

#SPJ4

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What is the pH of a solution made by mixing 200 mL of 0.025M HCl and 150 mL of 0.050 M HCl?
seraphim [82]

Answer:

The answer to your question is pH = 1.45  

Explanation:

Data

pH = ?

Volume 1 = 200 ml

[HCl] 1 = 0.025 M

Volume 2 = 150 ml

[HCl] 2 = 0.050 M

Process

1.- Calculate the number of moles of each solution

Solution 1

                Molarity = moles / volume

-Solve for moles

                moles = 0.025 x 0.2

result

               moles = 0.005

Solution 2

               moles = 0.050 x 0.15

-result

                moles = 0.0075

2.- Sum up the number of moles

Total moles = 0.005 + 0.0075

                   = 0.0125

3.- Sum up the volume

total volume = 200 + 150

                     350 ml or 0.35 l

4.- Calculate the final concentration

Molarity = 0.0125 / 0.35

              = 0.0357

5.- Calculate the pH

pH = -log [H⁺]

-Substitution

pH = -log[0.0357]

-Result

pH = 1.45    

8 0
3 years ago
Why does a cake rise when you bake it? ♡
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What are the monomers of bakelite​
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Answer:

Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.

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16 meters per hour to miles per sec
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Answer:

2.76 (full answer) 2.7617e-6

Explanation:

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3 years ago
A reaction has a rate constant of 1.15 x 10^−2 /s at 400K and 0.685 /s at 450K.
n200080 [17]

Answer:

a) the activation barrier = 122.3 kJ/mol

b) The rate constant at 425 K = 0.1001 /s

Explanation:

Step 1: Data given

Rate constant k1 = 1.15 * 10^−2 /s  at 400K (= T1)

Rate constant k2 = 0.685 /s at 450K (=T2)

Step 2: Determine the activation barrier for the reaction.

To determine the activation energy we will use the two-point Arrhenius equation:

ln(k₂/k₁) =  (Ea/R)((1/T1) - (1/T2))

⇒ with Ea = the activating energy

 ⇒ with R = the gas constant = 8.314 J/mol* K

⇒ with k1  = rate constant 1 = 1.15 *10^-2 /s

⇒ with T1 = Temperature 1 = 400 K

⇒ with k2 = rate constant 2 = 0.685/s

⇒ with T2 = temperature 2 = 450 K

= - (Ea/R)(T₁ - T₂)/T₁T₂

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

Ea = (8.314* ln(0.685/0.0115)) / ((1/400) - (1/450))

Ea = 122327.6 = 122.3 kJ/mol

B) What is the value of the rate constant at 425 K

For rate constant at 425 K.

Substitute the value of activation energy as 122327.6 J/mol, initial temperature as 400 K, final temperature as 425 K, rate constant at 400 K

1/T1   - 1/ T3   = 1/400   - 1 /425    = 1.47*10^-4

⇒ with T1 = the initial temperature = 400 K

⇒ with k1 = the rate constant at 400 K = 1.15 * 10^-2 /s

⇒ with T3 = the nex temperature = 425 K

⇒ with k3 = the rate constant at 425 K

ln(k3/k1) = Ea/R * ((1/T1)- (1/ T3))

⇒ with k3 = the rate constant at 425 K

⇒ with T3 = 425 K

k3/k1 = e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = k1* e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = 0.0115 * e^(122327.6/8.314 * (1.4710^-4))

k3 = 0.0115* e^2.1643

k3 = 0.1001 /s

4 0
3 years ago
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