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Vilka [71]
3 years ago
8

Sunlight Grass → Rabbit + Snake

Chemistry
2 answers:
Bezzdna [24]3 years ago
7 0

Answer:

The abiotic factor is <u>S</u><u>u</u><u>n</u><u>l</u><u>i</u><u>g</u><u>h</u><u>t</u><u>.</u>

Savatey [412]3 years ago
4 0
I think is sunlights I think that
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An unknown volume of gas has a pressure of 0.50 atm and temperature of 325 K. If the pressure is raised to 1.2 atm and the tempe
Grace [21]
You can solve this by using the equation (P1V1/T1) = (P2V2/T2). Plug in 0.50 atm for P1, leave V1 as the unknown, and plug in 325 K as T1. Then substitute 1.2 atm for P2, 48 L for V2, and 320 K for T2. Solve for V1, which is 117L, but since you round using two sig figs, your answer is C, 120 L. Hope this helps!
4 0
3 years ago
A+common+iv+solution+is+0.9%+saline+(nacl+solution).+what+is+the+osmolarity+of+0.9%+saline+mosmoles/l?+the+molecular+weight+of+n
Vedmedyk [2.9K]

An osmolarity of saline solution is 308 mosmol/L.

m(NaCl) = 9 g; the mass of sodium chloride

V(solution) = 1 L; the volume of the saline solution

n(NaCl) = 9 g ÷ 58.44 g/mol

n(NaCl) = 0.155 mol; the amount of sodium chloride

number of ions = 2

Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.

The osmolarity = n(NaCl) ÷ V(solution)  × 2

The osmolarity = 0.154 mol ÷ 1 L × 2

The osmolarity = 0.154 mol/L × 1000 mmol/m × 2

The osmolarity of the saline solution = 308 mosm/L.

More about osmolarity: brainly.com/question/13258879

#SPJ4

8 0
1 year ago
What chemical change is magnetizing steel
labwork [276]
Magnetizing steel is a physical change, not a chemical change.
4 0
3 years ago
Q3 You measure about 100 quarters...
tresset_1 [31]
The answer is four hundred
7 0
3 years ago
A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2
Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O (Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL

Putting values in above equation, we get:

2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL

M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M

0.823 M was the molarity of the KOH solution.

7 0
3 years ago
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