Calculate the mass in carbon units respectively: 7K , 12Si , 15P

Why are large numbers of animals found in areas where upwelling occurs

Ivan

Answer: this is because nutrients are increased by the movement of water.

Hope this helps! ^^

7 0

3 years ago

Should a mortar and pestle should be used for grinding only one substance at a time

MatroZZZ [7]

The mortar and pestle is most commonly used in chemical laboratories or in the kitchen. Its key function is to grind the material into smaller pieces, usually into its powdered form. It looks like that shown in the picture. The mortar is the bowl in which the material to be pounded is placed, and the pestle does the pounding.

Now, when you ask if it can only pound one at a time, my honest answer is, it depends. Depending on the size of your mortar, you could grind materials two or three at a time. But if you are concerned with contamination, then you do it one at a time, especially if you don't want them to get mixed up.

8 0

3 years ago

For water at 30C and 1 atm: a= 3.04x10^-4 k^-1 , k =4.52x10^-5 atm ^-1 = 4.46 x10^-10 m^2/N, cpm= 75.3j/9molk), Vm =18.1cm^3/mol

Lapatulllka [165]

Answer:

$C_{vm}$ of water at 30C and 1 atm is 256.834 J/mol·K.

Explanation:

To solve the question, we note the Maxwell relation such as

$C_{pm}-C_{vm}=\frac{9T\alpha ^2 V }{K}$

Where:

$C_{pm}$ = Specific heat of gas at constant pressure = 75.3 J/mol·K

$C_{vm}$ = Specific heat of gas at constant volume = Required

T = Temperature = 30 °C = 303.15 K

α = Linear expansion coefficient = 3.04 × 10⁻⁴ K⁻¹

K = Volume comprehensibility = 4.52 × 10⁻⁵ atm⁻¹

Therefore,

75.3 - $C_v$ = $\frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81 \times 10^{-5} }{4.52 \times 10^{-5} }$

$C_{vm}$ = $\frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81 \times 10^{-5} }{4.52 \times 10^{-5} }$ - 75.3 = 256.834 J/mol·K.

8 0

3 years ago

What causes the jet stream to move farther south?

Anni [7]

<span>Jet streams are the major means of transport for weather systems. A jet stream is an area of strong winds ranging from 120-250 mph that can be thousands of miles long, a couple of hundred miles across and a few miles deep. Jet streams usually sit at the boundary between the troposphere and the stratosphere at a level called the tropopause. This means most jet streams are about 6-9 miles off the ground. Figure A is a cross section of a jet stream.
</span>

The dynamics of jet streams are actually quite complicated, so this is a very simplified version of what creates jets. The basic idea that drives jet formation is this: a strong horizontal temperature contrast, like the one between the North Pole and the equator, causes a dramatic increase in horizontal wind speed with height. Therefore, a jet stream forms directly over the center of the strongest area of horizontal temperature difference, or the front. As a general rule, a strong front has a jet stream directly above it that is parallel to it. Figure B shows that jet streams are positioned just below the tropopause (the red lines) and above the fronts, in this case, the boundaries between two circulation cells carrying air of different temperatures.

7 0

3 years ago

Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h

Genrish500 [490]

Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

$k=\frac{0.693}{t_1_/_2}$

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

$k=\frac{0.693}{1.5 hrs}$

The expression for the rate constant is :

$k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}$

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

$\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6$

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

5 0

3 years ago