Answer
× 10²³ molecules are in 41.8 g of sulfuric acid
Explanation
The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.
Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol
Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.
Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.
Therefore, 0.426187053 moles of sulfuric acid is equal
Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.
Explanation:
Atomic number is defined as the number of an element which represents total number of protons.
When an atom is neutral then it means it contains same number of protons and electrons.
On the other hand, atomic mass is defined as the sum of total number of protons and neutrons present in an atom.
Protons of every element remains fixed because it shows the identity of each element but if we change the number of neutrons then also identity of the atom will remain fixed. This is because changing the number of neutrons will not show any change in number of protons.
For example, 
and 
are isotopes of hydrogen and they have same number of protons but different number of neutrons.
Thus, we can conclude that number of neutrons can vary without changing the identity of the element.
1.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element that reacts with water to produce a lilac flame
2.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element used as an inert atmosphere
3.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element that has a valency of 3
4.Write a balanced chemical equation for the reaction between potassium and water. (Non-anonymous question). .
(1 Point)
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5.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element with a fixed valency of 2 that not is not in group 2
Help
Complete Question:
This diagram shows a marble with a mass of 3.8 grams (g) that was placed into 10 milliliters (mL) of water. Using the formula V M D = , what is the density of the marble?
(See attachment for full diagram)
Answer:
1.27 g/cm³
Explanation:
First, find the volume of rock:
Volume of rock = volume of water after rock was placed - volume of water before rock was placed
Volume of rock = 13 - 10 = 3ml
Density of rock = grams of rock per 1 cm³
Note: 1 ml = 1 cm³
Let x represent amount of rock per 1 cm³
Thus,
3.8g = 3 cm³
x = 1 cm³
Cross multiply
1*3.8 = 3*x
3.8 = 3x
3.8/3 = 3x/3
1.27 = x
Density of rock = 1.27 g/cm³
