To find the ratio of the the combination for the ion, write the charge of the cation as the subscript for the anion, and the charge of the anion as the subscript of the cation. This will make the charges effectively cancel and you will be left with a neutral ionic compound. Remember, that an ionic compound is made up of a metal and a nonmetal.
For Ca2+ and Cl-, you will get the neutral compound to be CaCl₂.
Molar mass :
NaBr = 103 g/mol
Pb(NO3)2 = 331.20 g/mol
<span><span /><span>Balanced chemical equation :
</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
</span><span>
2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
g NaBr -------------------> 311 g Pb(NO3)2
331.20 g = 2*103*311
331.20 g = 64066
mass ( NaBr ) = 64066 / 331.20
mass ( naBr) = 193,43 g of NaBr
hope this helps!.
</span>
<h3>
Answer:</h3>
150000 J
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Thermodynamics</u>
Specific Heat Formula: q = mcΔT
- <em>q</em> is heat (in J)
- <em>m</em> is mass (in g)
- <em>c</em> is specific heat (in J/g °C)
- ΔT is change in temperature (in °C or K)
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] <em>m</em> = 225 g
[Given] <em>c</em> = 4.184 J/g °C
[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C
[Solve] <em>q</em>
<u>Step 2: Solve for </u><em><u>q</u></em>
- Substitute in variables [Specific Heat Formula]: q = (225 g)(4.184 J/g °C)(159.8 °C)
- Multiply: q = (941.4 J/°C)(159.8 °C)
- Multiply: q = 150436 J
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
150436 J ≈ 150000 J
Topic: AP Chemistry
Unit: Thermodynamics
Book: Pearson AP Chemistry
1.
The balanced chemical reaction is:
N2 +3 I2 = 2NI3
We are given the amount of product formed.
This will be the starting point of our calculations.
3.58 g NI3 ( 1 mol NI3 / 394.71 g NI3 ) ( 3
mol I2 / 2 mol NI3 ) = 0.014 mol I2.
Thus, 0.014 mol of I2 is needed to form the
given amount of NI3.