Answer:
58.44 g/mol
Explanation:
In this problem, make sure to remember that volume is measured in mL, L or any other units of volume. Remember that g represents grams, and grams is a measure of mass.
However, independent of what mass or what volume we take, molar mass is known to be an intensive property. That is, molar mass doesn't depend on any external conditions or any measurements.
Molar mass solely depends on the chemical structure of a compound and is a constant number at any given conditions.
In this problem, we are given sodium chloride, NaCl. In order to find its molar mass, we need to refer to the periodic table, find the atomic masses of Na and Cl and then add them up to have the molar mass of NaCl:

Answer:
490 in^3 = 8.03 L
Explanation:
Given:
The engine displacement = 490 in^3
= 490 in³
To determine the engine piston displacement in liters L;
(NOTE: Both in^3 (in³) and L are units of volume). Hence, to find the engine piston displacement in liters (L), we will convert in^3 to liters (L)
First, we will convert in³ to cm³
Since 1 in = 2.54 cm
∴ 1 in³ = 16.387 cm³
If 1 in³ = 16.387 cm³
Then 490 in³ = (490 in³ × 16.387 cm³) / 1 in³ = 8029.63 cm³
∴ 490 in³ = 8029.63 cm³
Now will convert cm³ to dm³
(NOTE: 1 L = 1 dm³)
1 cm = 1 × 10⁻² m = 1 × 10⁻¹ dm
∴ 1 cm³ = 1 × 10⁻⁶ m³ = 1 × 10⁻³ dm³
If 1 cm³ = 1 × 10⁻³ dm³
Then, 8029.63 cm³ = (8029.63 cm³ × 1 × 10⁻³ dm³) / 1 cm³ = 8.02963 dm³
≅ 8.03 dm³
∴ 8029.63 cm³ = 8.03 dm³
Hence, 490 in³ = 8029.63 cm³ = 8.03 dm³
Since 1L = 1 dm³
∴ 8.03 dm³ = 8.03 L
Hence, 490 in³ = 8.03 L
Answer:
The 2 ml and 2.0 ml is the same thing.
<span>For this reaction, oxidation number of Carbon in
CO would be +2 while oxidation number of carbon in CO2 would be +4 and so this
means that carbon has oxidized. Oxidation number of nitrogen in NO is +2. While
oxidation number of nitrogen in N2 is 0 so this means that nitrogen had reduced.
The reducing agent is the one which provides electrons by oxidizing itself so
in this case; CO is the reducing agent while the C in CO oxidized to produce
electrons. </span><span>I
am hoping that this answer has satisfied your query about and it will be able
to help you, and if you’d like, feel free to ask another question.</span>