Answer:
In the nucleus of an atom ,there are protons & neutrons. Protons have charge of 1.6× 10^-19 C, while neutrons have 0C charges. Electrons orbit outside the atom. Their charge is - 1.6 ×10^-19C
Explanation:
Answer:

Explanation:
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In this case, since the density is computed by dividing the mass of the substance by its occupied volume (d=m/V), we first need to realize that 0.8206 g/mL is the same to 0.8206 kg/L, which means we first need to compute the volume in L:

Then, solving for the mass in d=m/V, we get m=d*V and therefore the mass of gasoline in that full tank turns out:

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Answer:
0.0177 L of nitrogen will be produced
Explanation:
The decomposition reaction of sodium azide will be:

As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas
The molecular weight of sodium azide = 65 g/mol
The mass of sodium azide used = 100 g
The moles of sodium azide used = 
so 1.54 moles of sodium azide will give =
mol
the volume will be calculated using ideal gas equation
PV=nRT
Where
P = Pressure = 1.00 atm
V = ?
n = moles = 2.31 mol
R = 0.0821 L atm / mol K
T = 25 °C = 298.15 K
Volume = 
<h3>
Answer:</h3>
0.6 g NaCl
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)
[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂
<u>Step 2: Identify Conversions</u>
[RxN] Na₂CO₃ → 2NaCl
Molar Mass of Na - 22.99 g/mol
Molar Mass of C - 12.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol
Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
0.551373 g NaCl ≈ 0.6 g NaCl