CHEMISTRY HIGH SCHOOL

The usual units of density are :
cm3/ g
cm2/g
g/cm
Nm

Answers

Answer 1
Answer: The usual units of density are g/cm.

You may have also seen g/mL used for density. Keep in mind that 1 cm = 1 mL.
Answer 2
Answer:

g/cm^3

To answer directly its g/cm with the little three on top.

:)

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Determine the number of atoms in 25.0 grams of bromine, Br. (The mass of one mole of bromine is 79.90 g.)

Answers

Answer: 1,88×10²³ atoms of Br in 25g of Br

Explanation:

25g Br

1 mol of Br =79,9g Br

Number of Avogadro: 6,022×10²³ = 1 mol

25gBr× 1molBr/79,9gBr =

6,022×10²³ atoms of Br/1 mol Br =

1,88×10²³ atoms of Bromine

A scientist tests a new cough medicine by giving it to a group who have colds. The scientist gives another group with colds a liquid and tells them it is cough medicine. The people in both group are women between the age of 20 and 30 who normally are in good health

Answers

Here's your asnwer.

50

Answer:

posible mente es una equivocacion porque dijo que estaban refriados y les dió una medicina para la tos

An athlete with a high count of red blood cells will perform well in tests of physicalendurance because
(a) waste products will be removed from the body easily.
(b) excess levels of glucose will increase levels of respiration.
(c) the reduced plasma volume will lower carbon dioxide levels.
(d) high levels of oxygen will be available to the muscles.

Answers

Red blood cells are the carriers of oxygen from the lungs to everywhere else
in the body. If there is a high count of red cells in the blood, then high levels
of oxygen will be available to the muscles.

Find the grams of iron in 79.2 g of Fe2O3.

Answers

First you find the RFM (Relative Formula Mass) of Fe2O3
Fe2 = 56 x 2
= 112
O3 = 16 x 3
= 48

112 + 48 = 160

Moles = mass/RFM
           = 79.2/160
           =0.495

Then you rearrange the formula
Mass= moles x RFM
        = 0.495 x 112
        = 55.44 g
(To find mass of Iron have to use RFM of total iron in the substance example Fe2 ^^)

The atomic mass of titanium is 47.88 atomic mass units. This atomic mass represents the - explain please!

Answers

Okay, let's get this organized. 

Nucleus 
   ↓↓↓↓
Holds the Protons and Neutrons

Orbitals
   ↓↓↓↓
Electrons

Protons ⇒ Positive particle
Electrons ⇒ Negative charged particles 
Neutrons ⇒ No charge

Atomic number → Number of protons in the nucleus
Atomic mass → Protons + neutrons 

Answer: The atomic mass of titanium is 47.88 atomic mass units. This atomic mass represents the number of protons plus neutrons in the atom's nucleus. 

Suppose a 1.30 g nugget of pure gold has zero net charge. What would be its net charge after it has 1.68% of its electrons removed?

Answers

Answer:

The net charge of 1.3 g nugget of pure gold after 1.68% of its electrons are removed is 559 C

Explanation:

When an atom gains electrons it becomes negatively charged. Conversely, when it looses electrons the atoms becomes positively charged thus

To solve this question, we rely on the relationship between the nmber of particles present in a given mass of an atom, Avogadro's number and number of moles, n

The given variables are

mass of pure gold nugget = 1.30 g

Quantity of electrons removed = 1.68% of electrons present in the gold sample

Molar mass of gold = 197 g/mol

Avogadro's number = 6.02 × 10²³ atoms/mole  

qc = one electron charge = -1.06 × 10⁻¹⁹ C/electron

Electrical charge of gold nugget = 0 C

Number of electrons in one gold atom = 79 electrons

Solving for the number of prticles or gold atoms in 1.3 grams of gold we get

n mass/(molar mass) = 1.3/197 moles of gold =  0.0066 moles

number of particles in 0.0066 moles of gold N = n×N_(A) = 0.0066 × 6.02 × 10²³  = 3.97 × 10²¹ atoms

since 79 electrons are present per particle we have

3.97 × 10²¹ × 79 = 3.14 × 10²³ electrons

quantity of elecrtrons removed = 1.68% of  3.14 × 10²³ electrons =1.68/100 × 3.14 × 10²³ electrons = 0.0168 × 3.14 × 10²³ electrons = 5.3 × 10²¹ electrons

The net charge of 5.3 × 10²¹ electrons = 5.3 × 10²¹ electrons × -1.06 × 10⁻¹⁹ C/electron =

5.59 × 10² C = 559 C
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