Answer:
B
Explanation:
The debris of marine organisms such as corals, shells, and algae accumulate in the bottom of marine waters and accumulate with sediments. This debris is made up of calcium carbonate and over time the pressure and temperatures after sedimentation layers result to formation of limestone that transforms the calcium carbonate to the form of calcite.
Answer:
The answer to your question is: Molarity = 0.078
Explanation:
Data
HCl
V = 250 ml
T = 27°C = 300 °K
P = 141 mmHg = 0.185 atm
V2 = 70 ml
NaOH
V = 24.3 ml
Molarity NaOH = ?
Process
1.- Calculate the number of moles of HCl
PV = nRT
n = PV / RT
R = 0.082 atm l / mol K
n = (0.185)(0.25) / (0.082)(300)
n = 0.046 / 24.6
n = 0.0019 moles
2.- Calculate molarity of HCl
Molarity = moles / volume
Molarity = 0.0019 / 0.070
Molarity = 0.027
3.- Write the balanced equation
HCl + NaOH ⇒ H₂O + NaCl
Here, we observe that the proportion HCl to NaOH is 1:1 .
Then 0.0019 moles of HCl reacts with 0.0019 moles of NaOH.
4.- Calculate the molarity of NaOH.
Molarity = 0.0019 / 0.0243
Molarity = 0.078
<span>You can work out the following:
Mass = Density x Volume
Density = Mass Ă· Volume
Volume = Mass Ă· Density
The density of gold is 0.670205 oz/cm^3
The volume is length x width x depth = 7*3*2 = 42
The answer is Mass = 0.670205 x 42 = 281 ounces</span>
Answer:
Mass of C₂H₄N₂ produced = 3.64 g
Explanation:
The balanced chemical equation for the reaction is given below:
3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)
From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O
Molar masses of the compounds are given below below:
CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol
Comparing the mole ratios of the reacting masses;
CH₄ = 1.65/16 = 0.103
CO₂ = 13.5/44 = 0.307
NH₃ = 2.21/17 = 0.130
converting to whole number ratios by dividing with the smallest ratio
CH₄ = 0.103/0.103 = 1
CO₂ = 0.307/0.103 = 3
NH₃ = 0.130/0.103 = 1.3
Multiplying through with 5
CH₄ = 1 × 5 = 5
CO₂ = 3 × 5 = 15
NH₃ = 1.3 × 5 = 6.5
Therefore, the limiting reactant is NH₃
8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂
Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂
Mass of C₂H₄N₂ produced = 3.64 g