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olga_2 [115]
3 years ago
13

If the mass of a material is 104 grams and the volume of the material is 21 cm3, what would the density of the material be?

Chemistry
1 answer:
WITCHER [35]3 years ago
6 0
Remembering that
d = m ÷ v

d = ?
m = 104 g
v = 21 cm³

Therefore:

d = 104 ÷ 21

<span>d = 4,95 g÷cm³</span>
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All radioactive nuclides undergo what
kotykmax [81]
They undergo nuclear fission.
6 0
3 years ago
Explain how, the removal of thermal energy (heat) from matter affects its particles.
WINSTONCH [101]

Answer:

When a substance is heated, it gains thermal energy. Therefore, its particles move faster and its temperature rises. When a substance is cooled, it loses thermal energy, which causes its particles to move more slowly and its temperature to drop.

4 0
3 years ago
Read 2 more answers
Which of the following properties would be the same among isotopes of the same element?
Viefleur [7K]

Answer : The properties which would be the same among isotopes of the same element are Flammability and Color

Explanation :

Isotopes are the elements that have same number of electrons and protons but they differ in number of neutrons.

The chemical properties of an element are determined by its valence electrons and since isotope have same number of electrons, they have similar chemical properties.

But isotopes have different number of neutrons. Neutrons are the subatomic particles which are present in the nucleus. They are charge less particles and they have mass. Therefore when the elements have different number of neutrons, their atomic masses differ.

As a result, isotopes have different mass related physical properties.

Let us discuss the given properties now

1) Flammability : Flammability in simple words means the ability of the substance to catch fire or the ability to burn. This is mainly dependent on the reactivity which is in turn dependent on the number of electrons.

Since the isotopes have same number of electrons, they show similar flammability.

2) Density : Density is defined as mass per unit volume. This property involves mass and we know that isotopes have different masses. Therefore they have different densities too.

3) Melting point : Melting point is a physical property which is mainly dependent on the molecular weight of the substance. Since isotopes have different masses, they tend to show different melting points.

4) Color : The color of the element is mainly dependent on the arrangement of electrons around the nucleus. Isotopes have same number of electrons which means the arrangement would also be the same. Therefore the isotopes of the same element do not show different colors.

The properties which would be same among isotopes of the same element are

1) Flammability

4) Color

5 0
4 years ago
What is the molality of a solution containing 10.0 G of NA2 S04 and 1000.0 G of water
galina1969 [7]

Explanation:

1 literThe total of water is equal to 1000.0 g of water

we need to find the molality of a solution containing 10.0 g of dissolved in  Na₂S0₄1000.0 g of water

1. For that find the molar mass

Na:  2 x 22.99= 45.98

S:  32.07

O:  4 x 16= 64

The total molar mass is 142.05

We have to find the number of moles, y

To find the number of moles divide 10.0g by 142.05 g/mol.

So the number of moles is 0.0704 moles.

For the molarity, you need the number of moles divided by the volume. So, 0.0704 mol/1 L.

The molarity would end up being 0.0704 M

The molality of a solution containing 10.0 g of Na2SO4 dissolved in 1000.0 g of water is  0.0704 Mliter

7 0
2 years ago
The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat
guajiro [1.7K]
First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

                                             =  1 mol * 18 mol/g

                                            = 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

       = 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus 

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

     = 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g 

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

      = 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

                                                  = 1.881 KJ  +6.01 KJ + 4.514 KJ

                                                  = 12.405 KJ


5 0
3 years ago
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