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iragen [17]
2 years ago
12

Light ray 1 strikes the smooth surface, separating air and water as shown below. A horizontal line is drawn and the part above t

he line is labeled Air and the part below the line is labeled Water. There is one slanting ray labeled 1 which falls on the line at a point and there are four rays of light labeled 2, 3, 4, and 5 which move away from same point on the line. Ray 5 is exactly vertical to the line and below the line, ray 4 is on the right of ray 5 and makes an angle of about 25 degrees with ray 5. Ray 3 is above the line and makes an angle of about 25 degrees with the line. Ray 2 is on the right of ray 1 and ray 1 and ray 2 make angles of about 25 degrees with the vertical at the point where ray 1 strikes. Which of the following rays is the part of ray 1 that is reflected from the smooth surface? Ray 2 Ray 3 Ray 4 Ray 5
Chemistry
2 answers:
FinnZ [79.3K]2 years ago
8 0

I think the correct answer is ray 2.

Vera_Pavlovna [14]2 years ago
8 0

Answer:

it is ray two

Explanation:

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What is the pH of a solution after the addition of 30.0 mL of 0.100 M NaOH to 50.0 mL of 0.10 M HBr?
Pani-rosa [81]

Answer:

pH = 1.6

Explanation:

  • HBr + NaOH ⇒ NaBr + H2O

0 mL NaOH:

  • HBr ↔ H3O+  +  Br-

⇒ [ H3O+] = M HBr = 0.1 M

⇒pH = -log [H3O+] = 1

30 mL NaOH:

⇒ mol NaOH = 0.1 mol / L * 0.03 L =  3 E-3 mol

⇒ mol HBr = 0.05 L * 0.1 mol/L = 5 E-3 mol

⇒ M HBr = ( 5 E-3 mol - 3 E-3 mol) / 0.08 L = 0.025 M

⇒ pH = - log (0.025) = 1.6

7 0
3 years ago
1. What is the relationship between ecological succession and equlibrium in an ecosystem?
alexandr1967 [171]

Answer:

Equilibrium in ecology refers to a state that occurs such that a small disturbance or change is counter balanced by another change so that the community is restored to its original state. Thus, as a community goes through multiple changes through each stage of succession, it is not in equilibrium.

4 0
3 years ago
Density measurements were conducted on a 22.5oC sample of water which had a theoretical density of 0.997655 g/ml. A volume of 10
Sophie [7]

Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).

  • (1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:

p_{A} =9.98g/10.00mL=0.998g/mL\\p_{B} =15.61g/15.00mL=1.041g/mL\\p_{C} =25.65g/25.00mL=1.026g/mL

  • (2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):

p_{average}=\frac{p_{1}+p_{2}+p_{3}}{3}   \\p_{average}=\frac{(0.998+1.041+1.026)g/mL}{3}\\p_{average}=1.022g/mL

  • (3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:

%error=\frac{|p_{average}-p_{theoretical}|}{p_{theoretical}} *100

%error=\frac{|1.022g/mL-0.997655g/mL|}{0.997655g/mL}*100

%error=2.44 %

7 0
3 years ago
A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.
seropon [69]

<u>Answer:</u> The p-function of Zn^{2+} and NO_3^{-} ions are 2.51 and 2.14 respectively.

<u>Explanation:</u>

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

  • Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, M_1=(2\times 0.0031)=0.0062M

  • Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, M_2=(2\times 0.0042)=0.0084M

To calculate the concentration of nitrate ions in the solution, we use the equation:

M=\frac{M_1V_1+M_2V_2}{V_1+V_2}

Putting values in above equation, we get:

M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M

Calculating the p-function of zinc ions and nitrate ions in the solution:

  • <u>For zinc ions:</u>

\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]

\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51

  • <u>For nitrate ions:</u>

\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]

\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14

Hence, the p-function of Zn^{2+} and NO_3^{-} ions are 2.51 and 2.14 respectively.

5 0
3 years ago
How many moles of Al will be consumed when 0.400 mol of Al2O3 are produced in
Vladimir [108]

Answer:

C) 0.800 mol

Explanation:

  • 4Al + 3O₂ → 2Al₂O₃

In order to <u>convert from moles of Al₂O₃ into moles of Al</u>, we'll need to use<em> the stoichiometric coefficients</em>, using a conversion factor that has Al₂O₃ moles in the denominator and Al moles in the numerator:

  • 0.400 mol Al₂O₃ * \frac{4molAl}{2molAl_2O_3} = 0.800 mol Al

So the correct answer is option C).

7 0
2 years ago
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