Answer:
Technically a microorganism or microbe is an organism that is microscopic. The study of microorganisms is called microbiology. Microorganisms can be bacteria, fungi, archaea or protists. The term microorganisms does not include viruses and prions, which are generally classified as non-living.
The number of hours required : 37.2 hours
<h3>Further explanation</h3>
Given
⁴²K (potassium -42)
Required
The number of hours
Solution
The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually, radioactive elements have an unstable atomic nucleus.
Based on Table N(attached), the half-life for ⁴²K is 12.4 hours, which means half of a sample of ⁴²K will decay in 12.4 hours
For three half-life periods :
![\tt 3\times 12.4=37.2~hours](https://tex.z-dn.net/?f=%5Ctt%203%5Ctimes%2012.4%3D37.2~hours)
The rover was designed to explore A.Mars
Answer:
![\Delta \rho =2.22 g/mL](https://tex.z-dn.net/?f=%5CDelta%20%5Crho%20%3D2.22%20g%2FmL)
Explanation:
Hello,
In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:
![\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL](https://tex.z-dn.net/?f=%5CDelta%20%5Crho%3D%5Crho%20_f-%5Crho%20_i%5C%5C%20%5C%5C%5CDelta%20%5Crho%3D5.43g%2FmL-3.21g%2FmL%5C%5C%5C%5C%5CDelta%20%5Crho%20%3D2.22%20g%2FmL)
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Answer:
Compound B has greater molar mass.
Explanation:
The depression in freezing point is given by ;
..[1]
![m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BMass%20of%20solvent%20in%20kg%7D%7D)
Where:
i = van't Hoff factor
= Molal depression constant
m = molality of the solution
According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.
The depression in freezing point of solution with A solute: ![\Delta T_{f,A}](https://tex.z-dn.net/?f=%5CDelta%20T_%7Bf%2CA%7D)
Molar mass of A = ![M_A](https://tex.z-dn.net/?f=M_A)
The depression in freezing point of solution with B solute: ![\Delta T_{f,B}](https://tex.z-dn.net/?f=%5CDelta%20T_%7Bf%2CB%7D)
Molar mass of B = ![M_B](https://tex.z-dn.net/?f=M_B)
![\Delta T_{f,A}>\Delta T_{f,B}](https://tex.z-dn.net/?f=%5CDelta%20T_%7Bf%2CA%7D%3E%5CDelta%20T_%7Bf%2CB%7D)
As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.
![\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}](https://tex.z-dn.net/?f=%5CDelta%20T_f%5Cpropto%20%5Cfrac%7B1%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%7D)
![M_A](https://tex.z-dn.net/?f=M_A%3CM_B)
This means compound B has greater molar mass than compound A,