Ask question

Ask question

All categories

3 years ago

8

What is true about constructive interference?A.Amplitudes in the opposite direction interfere to make larger waves.B.Amplitudes

in the same direction interference to larger waves.C.Amplitudes in the same direction interference to make smaller waves.

1 answer:

Alecsey [184]3 years ago

6 0

Answer:

B. Amplitudes in the same direction interference to larger waves

Explanation:

This is because constructive interference is the addition of waves to form a larger wave. In constructive interference, waves with amplitudes moving in the same direction add to form a larger wave.

You might be interested in

Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

<u>(a) 1.457 mol of KCl in 1.500 L of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

<u>(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

<u>(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

<u>(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

<u>(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

<u>(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago

If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN

viktelen [127]

Answer : The concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is $SCN^-$ and $Fe^{3+}$ is excess reagent.

First we have to calculate the moles of KSCN.

$\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}$

$\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol$

Moles of KSCN = Moles of $K^+$ = Moles of $SCN^-$ = $1.08\times 10^{-5}mol$

Now we have to calculate the concentration of $[Fe(SCN)]^{2+}$

$\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}$

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

$\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M$

Thus, the concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

7 0

3 years ago

B vitamins are a group of water soluble compounds that are important in cell metabolism. Each one has a different chemical struc

Tom [10]

Molar mass of vitamin B1, C12H17N4OS = 265.34 g/ mol

Molar mass of vitamin B2, C17H20N4O6 = 376.37 g/ mol

Molar mass of vitamin B5, C9H17NO5 = 219.24 g/ mol

Molar mass of vitamin B6, C8H11NO3 = 169.18 g/ mol

Molar mass of vitamin B7, C10H16N2O3S = 244.31 g/ mol

Now,

Order of increasing molar mass = B6 < B5 < B7 < B1 < B2

7 0

4 years ago

Using distance and can help you identify the position of something

miskamm [114]

Answer: Length and width

Explanation:

4 0

3 years ago

Read 2 more answers

In the image above, the region labeled “ A “ is called the . . .

tatyana61 [14]

The awnser is central canal the third one

8 0

2 years ago

Read 2 more answers