Answer:
Mass = 51 g
Explanation:
Given data:
Mass of nitrogen = 41.93 g
Mass of ammonia formed = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 41.93 g/ 28 g/mol
Number of moles = 1.5 mol
now we will compare the moles of nitrogen and ammonia.
N₂ : NH₃
1 : 2
1.5 : 2/1×1.5 = 3 mol
Mass of ammonia formed:
Mass = number of moles × molar mass
Mass = 3 mol × 17 g/mol
Mass = 51 g
Answer:
Metamorphic rock is classified by texture and composition. The texture of a metamorphic rock can be either foliated and appear layered or banded, or non-foliated and appear uniform in texture without banding. Foliated rocks contain many different kinds of minerals, but non-foliated rocks contain only one main mineral, which contributes to their more uniform appearance. Igneous rocks are classified according to mode of occurrence, texture, mineralogy, chemical composition, and the geometry of the igneous body.
Explanation:
Answer:the CO2 molecule has an excess of electron
Answer:
0.03g/mL
Explanation:
Given parameters include:
Five μL of a 10-to-1 dilution of a sample; This implies the Volume of dilute sample is given as 5 μL
Dilution factor = 10-to-1
The absorbance at 595 nm was 0.78
Mass of the diluted sample = 0.015 mg
We need to first determine the concentration of the diluted sample which is required in calculating the protein concentration of the original solution.
So, to determine the concentration of the diluted sample, we have:
concentration of diluted sample = 
=
(where ∝ was use in place of μ in the expressed fraction)
= 0.003 mg/μL
The dilution of the sample is from 10-to-1 indicating that the original concentration is ten times higher; as such the protein concentration of the original solution can be calculated as:
protein concentration of the original solution = 10 × concentration of the diluted sample.
= 10 × 0.003 mg/μL
= 0.03 mg/μL

= 0.03g/mL
Hence, the protein concentration of the original solution is known to be 0.03g/mL
Energy of the reactants would be correct