<h3>
Answer:</h3>
2.999 mol Br
<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>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
1.806 × 10²⁴ molecules Br
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
<u />
= 2.999 mol Br
<u>Step 4: Check</u>
<em>We are given 4 sig figs. Follow sig fig rules and round.</em>
Our final answer is already in 4 sig figs, so there is no need to round.
Answer:
Equation of Reaction
2AgNO3 + BaCl2 === 2AgCl + Ba(NO3)2
Molar Mass of AgNO3 = 170g/mol
Moles of reacting AgNO3 = 100g/170gmol-¹
=0.588moles of AgNO3
From the equation of reaction...2moles of AgNO3 reacts to Produce 2Moles of Silver Chloride
So Their ratio is 2:2.
This means that 0.588Moles of AgCl Will be produced too.
ANSWER...0.588MOLES OF AgCl WILL BE PRODUCED.
Answer:
Uh first of all this is algebra but I'll answer this
First distribute the three and 5 (Multiply them by both terms inside parenthesis.
3x-6=5x+20
Then add like terms
8x=14
Divide 8 by 8 and 8 by 14
x = 14/8
Explanation:
Answer:
Explanation:
Given parameters:
number of moles magnesium = 6.80mol
number of moles of chlorine = 13.56mol
To find the complete chemical formula, we should obtain the formula of the compound.
Mg Cl
Number of
moles 6.8 13.56
Dividing
by the smallest 6.8/6.8 13.56/6.8
1 2
The formula of the compound is MgCl₂
This is an ionic compound in which Magnesium loses two electrons that would be gained by Cl atoms requiring just an electron each to complete their octet.
C5H12 (l) + 8O2 (g) ----> 5CO2 (g) + 6H2O (l)
Delta H = -3505.8 kJ/mol
C (s) + O2 (g) -----> CO2 (g)
Delta H = -393.5 kJ/mol
H2 (g) + (1/2)O2 (g) ------> H2O (l)
Delta H = -286 kJ/mol
Possible answers:
a. +35 kJ/mol
b. + 1,073 kJ/mol
c. -4,185 kJ/mol
d. -2,826 kJ/mol
e. -178 kJ/mol
