<u>Answer:</u>
The correct answer option is C. 2.
<u>Explanation:</u>
We are given the number '0.0020' and we are to indicate the number of significant figures in the given measured number.
According to the rules of significant figures, numbers that are non-zero, zeros between any two significant numbers and the ending zeros in the decimal position are categorized as significant figures.
Since there is one non-zero number and one ending zero in the decimal position, therefore 0.0020 has 2 significant figures.
Answer:
22Ω
Explanation:
Given parameters:
Potential difference = 3.3V
Current = 0.15A
Unknown:
Resistance = ?
Solution:
According to ohm's law, potential difference, current and resistance are related by the expression below;
V = I R
where V is the voltage
I is the current
R is the resistance
3.3 = 0.15 x R
R = 
= 22Ω
Answer:
density=1.43 g/L
Explanation:
Since the density formula is density = mass / volume, we need to find out the mass of the gas and the volume is that of the container.
The mass of the gas is 130.0318 g-129.6375 g=0.3943 g
The gas volume is 276mL*(1L/1000mL) 0.276 L
density = mass / volume=0.3943g/0.276L
density =1.43g/L
Answer:
Acids change the color of litmus from blue to red.
They convert the color of Methyl Orange from Orange/Yellow to Pink.
Acids turn the pink color of Phenolphthalein to colourless.
Acids can conduct electricity.
Some Acids are highly corrosive in nature which means that they corrode or rust metals.
Explanation:
hope this helped
ur boi johnny joestar.
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
