Answer:
a) 2.01 g
Explanation:
- Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃
First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:
- 0.0302 mol AgNO₃ *
= 0.0151 mol Na₂CO₃
So the remaining Na₂CO₃ moles are:
- 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃
Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:
- 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃
The closest answer is option a).
5H2O2 + 2KMnO4<span>+ 3H2SO4 = 5O2 + 2MnSO4 + 8H2O + K2SO4
0,145 moles of KMnO4----------in--------1000ml
x moles of KMnO4---------------in------------46ml
x = 0,00667 moles of KMnO4
according to the reaction:
2 moles of KMnO4------------------5 moles of H2O2
0,00667 moles of KMnO4----------------x
x = 0,01668 moles of H2O2
0,01668 moles of H2O2---------in-----------50ml
x moles of H2O2--------------------in----------1000ml
<u>x = 0,334 mol/L H2O2</u></span>
Answer:
2 ATP molecules are produced
Answer and explanation:
The attached figure shows five different structures for the chemical formula C4H5O5, but only one of these structures represent the malic acid.
Malic acid is a dicarboxylic acid, this means malic acid has two
-COOH groups. Also, malic acid is a secundary alcohol, which means it has a R2-C-OH group.
-Structure A has two carboxylic groups, but it doesn´t have a secundary alcohol.
-Structure B doesn´t have any caborxylic group.
-Structure C has two carboxylic groups and it is a secundary alcohol. Structure C is the Malic acid
Answer:
3.2 L
Explanation:
Given data:
Mass of oxygen = 3.760 g
Pressure of gas = 88.4 Kpa (88.4×1000 = 88400 Nm⁻²)
Temperature = 19°C (19+273.15 = 292.15 K)
R = 8.314 Nm K⁻¹ mol⁻¹
Volume occupied = ?
Solution:
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 3.760 g/ 32 g/mol
Number of moles = 0.12 mol
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant
T = temperature in kelvin
V = nRT/P
V = 0.12 mol × 8.314 Nm K⁻¹ mol⁻¹ × 292.15 K /88400 Nm⁻²
V = 291.472 Nm /88400 Nm⁻²
V = 0.0032 m³
m³ to L:
V = 0.0032×1000 = 3.2 L
