Waves we cannot actually see (unlike ripples) and those not needing a medium to travel within belong to this category of waves?

In a constant-pressure calorimetry experiment, a reaction gives off 22.8 kJ of heat. The calorimeter contains 150 g of water, in

ludmilkaskok [199]

Answer:

The final temperature of water is 56,7 ºC

Explanation:

We need to apply the specific heat capacity formula to solve this.

Q = m . C . ΔT

22,8 kJ = 150 g . 4,186 J/g °C . (Tfinal - 20,4º C)

(We need to convert kJ in J to use the specific heat of water, be careful with units)

22,8 kJ = 22800 J (x1000)

22800 J = 150 g . 4,186 J/g °C . (Tfinal - 20,4º C)

22800 J = 627,9 J/ºC (Tfinal - 20,4º C)

22800 J / 627,9 ºC/ J = (Tfinal - 20,4º C)

36,3ºC = Tfinal - 20,4º C

36,3ºC + 20,4º C = 56,7 ºC

7 0

4 years ago

How many ml of a 14. 0 m nh3 stock solution are needed to prepare 200 ml of a 4. 20 m dilute nh3 solution?.

Lilit [14]

Answer:

60 mL

Explanation:

200 mL = 0.2 L

We know that:

Molarity = Moles / Liters

So:

4.20 M = moles / 0.2 L

moles = 4.20 x 0.2 = 0.84

Then:

14.0 M = 0.84 moles / Liters

Liters = 0.84 moles / 14.0 M = 0.06 L

Since they want the answer in mL, convert 0.06 to get 60 mL

5 0

2 years ago

How many moles of glucose, C6H12O6, can be burned when 60.0 mol of oxygen is available? C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

kiruha [24]

60mol O2 × 1 mol C6H12O6 / 6 moles 02 = 10 moles of Glucose

6 0

4 years ago

(A) 2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(l) (B) C2H3O2-(aq) + H3O+(aq) → HC2H3O2(aq) + H2O(l) (C) 4 H+(aq) + 4 Co2+(aq) + O2

Anestetic [448]

Answer:

The reaction (E) 2 H₂O₂ (l) → O₂ g) + 2 H₂O (l) has a single species that is both oxidized and reduced.

Explanation:

Given the reaction (E):

2 H₂O₂ (l) → O₂ g) + 2 H₂O (l)

In hydrogen peroxide, oxygen has an oxidation state of -1. We can see that the reaction is balanced.

The products of the reaction are:

O₂ (the oxidation state of the Oxygen is 0)

H₂O (the oxidation state of the Oxygen is -2)

Hence, 1 mol of Oxygen changes its oxidation state from -1 to 0 (oxidation) and 1 mol of Oxygen changes its oxidation state from -1 to -2 (reduction).

The reaction (E) has a single species that is both oxidized and reduced.

6 0

3 years ago

Base your answer on the information below. The radioisotope uranium-238 occurs naturally in Earth's crust. The disintegration of

algol [13]

Answer:

12 mmilligrams of Po-218 was the mass of the original starting material

Explanation:

The half-life of a radioactive material is the time taken for half the amount ofnthe original material present in a radioactive material to decay or disintegrate.

After each half-life, half the original material present at the start remains.

For the radioactive polonium-218 having a half-life of 3.04 minutes, it means that if 1 g is the starting material, after 3.04 minutes, 1/2 g will be remaining; after, 6.08 minutes 1/2 of 1/2 which is 1/4 of the starting material will be remaining; and after 9.12 minutes, 1/2 of 1/4 = 1/8 g will be remaining.

From the question, number of half-lives undergone after 9.12 minutes = 9.12/3.04 = 3 half-lives.

After 3 half-lives, 1/8 of the original material is remaining.

1/8 = 1.50 mg

The original mass of the sample at the start = 1.50 mg × 8 = 12 mg

Therefore, 12 milligrams of Po-218 was the mass of the original starting material.

4 0

3 years ago