How many moles of water molecules are in 57.43 g of iron (iii) chloride dihydrate

tate whether the following changes are physical or chemical for rancidipication fixation of water 2 tearing of paper 3 rusting o

damaskus [11]

Answer: Physical change : tearing of paper, fixing of wtaer

Chemical change: rusting of iron , electrolysis of water, Rancidification

Explanation:

Physical change is a change in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.

Example: tearing of paper, fixing of wtaer

Chemical change is a change in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.

Example: rusting of iron , electrolysis of water, Rancidification

3 0

3 years ago

There are 50 fish in the pond that have tags. You catch 25 fish, and 5 of them have tags. What is the best estimate of the numbe

sergeinik [125]

Answer: 250

Explanation:

You work this problem by using proportions.

A proportion is the equalization of two ratios.

Here you assume that the ratio of fish with tags to total fish that you catch is the same than the ratio of fish with tags to total fish in the pond.

Mathematically:

5 fish with tag / 25 fish = 50 fish with tag / x

5 / 25 = 50 / x

Solve for x:

Multiplication property of equality: x × 5 = 50 × 25
Division property of equality: x = 50 × 25 / 5
Result: 250

8 0

3 years ago

Read 2 more answers

A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final

VMariaS [17]

M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C

m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C

T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C

7 0

4 years ago

What is the binding energy of a nucleus that has a mass defect of 5.81*10-^29 kg

IrinaVladis [17]

Answer:

Choice A: Approximately $5.23 \times 10^{-29}$ joules.

Explanation:

Apply the famous mass-energy equivalence equation to find the energy that correspond to the $\rm 5.81\times 10^{-29}$ kilograms of mass.

$E = m \cdot c^{2}$ ,

where

$E$ stands for energy,
$m$ stands for mass, and
$c$ is the speed of light in vacuum.

The speed of light in vacuum is a constant. However, finding the right units for this value can simplify the calculations a lot. What should be the unit of $c$ ?

The mass given is in the appropriate SI unit:

Mass is in kilograms.

Thus, proceed with the speed of light in SI units. The SI unit for speed is meters per second. For the speed of light, $c \approx \rm 3.00\times 10^{8}\;m\cdot s^{-1}$ .

Apply the mass-energy equivalence:

$\begin{aligned} E &= m \cdot c^{2} \\ &= \rm 5.81\times 10^{-29}\; kg \times {\left(3.00\times 10^{8}\; m\cdot s^{-1}\right)}^{2}\\ &\approx \rm 5.23\times 10^{-12}\;kg\cdot m^{2}\cdot s^{-2} \end{aligned}$ .

The unit of energy is not in joules. Don't be alerted. Consider the definition of a joule of energy. One joule is the work done on an object when a force of one newton acts on the object in the direction of the force through the distance of one meter. (English Wikipedia.)

$\rm 1\; J = 1\; N \times 1\; m$ .

However, a force of one newton is defined as the force required to accelerated an object with a mass of one kilogram (not gram) at a rate of one meter per second squared. (English Wikipedia.)

$\begin{aligned}\rm 1\; J &= \rm 1\; N \times 1\; m\\ & = \rm \left(1\; kg\times 1\; m\cdot s^{-2}\right)\times 1\; m\\ &= \rm 1\; kg \cdot m^{2}\cdot s^{-2}\end{aligned}$ .

In other words, the mass defect here is also $\rm 5.23\times 10^{-12}\; J$ .

7 0

3 years ago

Read 2 more answers

The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2

Gre4nikov [31]

Answer:

No the substance is not water.

Explanation:

The balance chemical equation for the decomposition of water is as follow;

2 H₂O = 2 H₂ + O₂

Step 1: Calculate moles of H₂O;

Moles = Mass / M.Mass

Moles = 5.0 g / 18.01 g/mol

Moles = 0.277 moles of H₂O

Step 2: Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:

According to equation,

2 moles of H₂O produced = 1 mole of O₂

So,

0.277 moles of H₂O will produce = X moles of O₂

Solving for X,

X = 0.277 mol × 1 mol / 2 mol

X = 0.138 moles of O₂

Also,

According to equation,

2 moles of H₂O produced = 2 mole of H₂

So,

0.277 moles of H₂O will produce = X moles of H₂

Solving for X,

X = 0.277 mol × 2 mol / 2 mol

X = 0.227 moles of H₂

Step 3: Calculate Mass of O₂ and H₂ as;

For O₂:

Mass = Moles × M.Mass

Mass = 0.138 mol × 31.99 g/mol

Mass = 4.44 g of O₂

For H₂:

Mass = Moles × M.Mass

Mass = 0.227 mol × 2.01 g/mol

Mass = 0.559 g of H₂

Conclusion:

From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

6 0

3 years ago