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3 years ago

7

suppose you want to do an experiment to hear how three different objects would sound when you dropping them the same distance. y

ou drop a pebble a rock,and a grain of sand.what are the variables of your experiment

1 answer:

hodyreva [135]3 years ago

8 0

The variable would be the distance you dropped them.

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The individual particles in an ionic solid are held together as a result of:

dlinn [17]

electrostatic force

Explanation:

In ionic bonds one atom is positively charged (the atom that gives its electrons), and the other atom is negatively charged (the atom that receives the electrons); therefore, it is the opposite charge of the ions that attracts them to one another and holds them together with electrostatic force.

8 0

2 years ago

A compound containing sodium, chlorine, and oxygen is 25.42% sodium by mass. A 3.25 g sample gives 4.33×1022 atoms of oxygen. Wh

Tcecarenko [31]

step one
calculate the % of oxygen
from avogadro constant
1moles = 6.02 x 10 ^23 atoms
what about 4.33 x10^22 atoms
= ( 4.33 x 10^ 22 x 1 mole ) / 6.02 10^23= 0.0719 moles
mass= 0.0719 x16= 1.1504 g
% composition is therefore= ( 1.1504/3.25) x100 = 35.40%
step two
calculate the % composition of chrorine
100- (25.42 + 35.40)=39.18%

step 3
calculate the moles of each element
that is
Na = 25.42 /23=1.1052 moles
Cl= 39.18 /35.5=1.1037moles
O= 35.40/16= 2.2125 moles
step 4
find the mole ratio by dividing each mole by 1.1037 moles
that is
Na = 1.1052/1.1037=1.001
Cl= 1.1037/1.1037= 1
0=2.2125 = 2
therefore the empirical formula= NaClO2

8 0

3 years ago

A silver cube with an edge length of 2.38 cm and a gold cube with an edge length of 2.79 cm are both heated to 81.9 ∘C and place

never [62]

Answer:

The final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

Explanation:

<u>Given data;</u>

edge length of silver, a = 2.38 cm = 0.0238 m

edge length of gold, a = 2.79 cm = 0.0279 m

final temperature of silver, t = 81.9 ° C

final temperature of Gold, t = 81.9 ° C

initial temperature of water, t = 19.6 ° C

volume of water, v = 109.5 mL = 0.0001095 m³

<u>Known data:</u>

density gold 19300 kg/m³

density silver 10490 kg/m³

density water 1000 kg/m³

specific heat gold is 129 J/kgC

specific heat silver is 240 J/kgC

specific heat water is 4200 J/kgC

<u>Calculated data</u>

Apply Pythagoras theorem to determine the side of each cube;

Silver cube;

let L be the side of the silver cube

Taking the cross section of the cube (form a right angled triangle), the edge length forms the <em>hypotenuse side</em>.

L² + L² = 0.0238²

2L² = 0.0238²

L² = 0.0238² / 2

L² = 0.00028322

L = √0.00028322

L = 0.0168

Volume of cube = L³

Volume of the silver cube = (0.0168)³ = 4.742 x 10⁻⁶ m³

Gold cube;

let L be the side of the gold cube

L² + L² = 0.0279²

2L² = 0.0279²

L² = 0.0279² / 2

L² = 0.0003892

L = √0.0003892

L = 0.0197

Volume of cube = L³

Volume of the silver cube = (0.0197)³ = 7.645 x 10⁻⁶ m³

Mass of silver cube;

density = mass / volume

mass = density x volume

mass of silver cube = 10490 (kg/m³) x 4.742 x 10⁻⁶ (m³) = 0.0497 kg

Mass of Gold cube

mass of gold cube = 19300 (kg/m³) x 7.645 x 10⁻⁶ (m³) = 0.148 kg

Mass of water

mass of water = 1000 (kg/m³) x 0.0001095 (m³) = 0.1095 kg

Let the heat gained by cold water be Q₁

Let the heat lost by silver cube = Q₂

Let the heat lost by gold cube = Q₃

Let the final temperature of water = T

Q₁ = 0.1095 kg x 4200 J/kgC x (T–19.6)

Q₂ = 0.0497 kg x 240 J/kgC x (81.9–T)

Q₃ = 0.148 kg x 129 J/kgC x (81.9–T)

At thermal equilibrium;

Q₁ = Q₂ + Q₃

0.1095 x 4200 (T–19.6) = 0.0497 x 240 x (81.9–T) + 0.148 x 129 x (81.9–T)

459.9 (T–19.6) = 11.928 (81.9–T) + 19.092(81.9–T)

459.9T - 9014.04 = 976.9032 - 11.928T + 1563.6348 - 19.092T

459.9T + 11.928T + 19.092T = 9014.04 + 976.9032 + 1563.6348

490.92T = 11554.578

T = 11554.578 / 490.92

T = 23.54 ⁰C

Therefore, the final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

6 0

3 years ago

In a synthesis reaction, one reactant contains 346 J of chemical energy, and one reactant contains 153 J of chemical energy. The

Eduardwww [97]

Answer:

64J of energy must have been released.

Explanation:

Step 1: Data given

One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.

The product contains 435 J of chemical energy.

Step 2:

Since the energy is conserved

Sum of energy of Reactants = Energy of Products

Sum of energy of Reactants = 346 J + 153 J = 499 J

The energy of the product = 435 J

435 < 499

This means energy must have been lost as heat.

Step 3: Calculate heat released

499 J - 435 J = 64 J

64J of energy must have been released.

4 0

3 years ago

When water reaches the metastable state, the phase of water will be?

Jet001 [13]

Answer:

water, when the metastable state is reached, is cooled below the zero temperature. It freezes abruptly. this is called metastable. They are not at equilibrium per se; as at negative temperatures the only equilibrium state of water is ice.

Explanation:

7 0

3 years ago