Which subatomic particles are found in the

What is work and state the law of conservation mass?

Butoxors [25]

Answer:

Work is a force causing the movement or displacement of an object

law of conservation mass:

1. Atoms cannot be created or destroyed in a chemical reaction.

2. Molecules cannot be created or destroyed in a chemical reaction.

3. Compounds cannot be created or destroyed in a chemical reaction.

4. Heat cannot be created or destroyed in a chemical reaction.

7 0

3 years ago

How many moles of sodium hydroxide are there in 1.0mL of 2.0M NaOH

kvasek [131]

To find the moles, you can use the following formula

moles= Molarity x Liters

Molarity= 2.0 M
Liters= 0.0010 Liters ---------------->>>>>>>>>> 1.0 mL= 0.0010 Liters

moles= 2.0 M x 0.0010 Liters= 0.0020 moles

4 0

2 years ago

a compound has 15.39 g of gold for every 2.77 g of chlorine. simplified there is _____ g of gold for every 1 g of chlorine

iren [92.7K]

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction $\frac{a}{b}$

You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:

$\frac{15.39 g of gold}{2.77 g of chlorine}$

The proportion is the equal relationship that exists between two reasons and is represented by: $\frac{a}{b}=\frac{c}{d}$

This reads a is a b as c is a d.

To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:

$\frac{15.39 g of gold}{2.77 g of chlorine}=\frac{mass of gold}{1 g of chlorine}$

Solving for the mass of gold gives:

$mass of gold=1 g of chlorine*\frac{15.39 g of gold}{2.77 g of chlorine}$

mass of gold= 5.56 grams

So, there is 5.56 g of gold for every 1 g of chlorine

5 0

2 years ago

Calcium bicarbonate (Ca(HCO3)2) Ca:H:C:O = 1:2:2 ___ Lithium sulfide (Li:2s) Li:S = 2:__

Anastaziya [24]

Answer: The ratio of atoms in calcium bicarbonate ; Ca : H : C : O = 1:2:2:6.

The ratio of atoms in lithium sulfide; Li : S = 2 : 1

Explanation:

In calcium bicarbonate: $Ca(HCO_3)_2$

In a molecular formula of calcium carbonate there are:

Number of Calcium atoms = 1

Number of Hydrogen atom = 1 × 2 = 2

Number of Carbon atoms = 1 × 2 = 2

Number of Oxygen atoms = 3 × 2 = 6

So, Ca : H : C : O = 1 : 2 : 2 : 6

In lithium sulfide : $Li_2S$

In a molecular formula of lithium sulfide there are:

Number of Lithium atoms = 1 × 2 = 2

Number of Sulfur atoms = 1

So, the Li : S = 2 : 1

8 0

2 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0

3 years ago