What role does the grass play in the food web?

nataly862011 [7]

First you need to know how many isotopes there are of silicon, and its average atomic units (look at periodic table). Then make up a system of equations to solve for it. Theres 3 stable silicon isotopes (28, 29, 30) so you will need to have 3 equations. You must be given the percent abundance of at least one of the isotopes to solve because here I can only see 2 equations (numbered down below) set x = percent abundance of si-28 y = percent abundance of si-29 z = percent abundance of si-30 since all of silicon atoms account for 100% of all silicon: x + y + z = 100% = 1 therefore: 1) x = 1 - y - z You also have 2) 28x + 29y + 30z = average atomic mass you can substitute x so that equation becomes: 28 (1 - y - z) + 29y + 30z = average atomic mass See how you have 2 variables here? You cant go on until you know the value of one isotope already or you have given a clue which you can derive the third equation

5 0

3 years ago

Is water made of plant cells or animal cells •plant •animals •niether

Pepsi [2]

Answer:

i think niether

Explanation:

5 0

2 years ago

Which of the following is true about an atom of silver

kati45 [8]

Answer:

where is options.

tell me

4 0

3 years ago

Read 2 more answers

Convert 11.03 moles of calcium nitrate to grams

hichkok12 [17]

Im pretty sure the answer is 0.01218859659g

not 100% sure tho so please consult someone else b4 answering
i hope this helps!!

6 0

3 years ago

Read 2 more answers

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago