It would be +3
Since the number of protons and electrons must be equal in an atom for it to be neutral, if it looses electrons it becomes positively charged ion and if it gains electrons it becomes negatively charged ion.
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
Explanation:
“This new data set shows that as surface temperature increases, so does atmospheric humidity,” Dessler said. “Dumping greenhouse gases into the atmosphere makes the atmosphere more humid. And since water vapor is itself a greenhouse gas, the increase in humidity amplifies the warming from carbon dioxide."
Answer:
670.68°C
Explanation:
Given that:
volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g
specific heat (C) = 4.184 J/g˚C
Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C
The quantity of heat (Q) used to raise the temperature of a body is given by the equation:
Q = mCΔT
Substituting values:
Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J
Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.
-Q = mCΔT
-418.4 J = 5 g × 0.129 J/g˚C × ΔT
ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C
temperature change ΔT = final temperature - initial temperature
- 648 .68°C = 22°C - Initial Temperature
Initial Temperature = 22 + 648.68 = 670.68°C
Answer;
294.13 amu
Solution;
-293nv is 293.10 amu and that of 295nv is 295.45 amu
293.05+295.2=588.25
588.25/2= 294.13 amu
an amu of 294.13 using significant figures
