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2 years ago

Which waves carries the most energy?

Chemistry

2 answers:

Vadim26 [7]2 years ago

4 0

Answer:

I think c

Explanation:

marin [14]2 years ago

4 0

It’s definitely choice C.

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Which of the following reactions is a neutralization reaction? A. ZnCl2(aq) + CaCrO4(aq) → ZnCrO4(s) + CaCl2(aq) B. HNO3(aq) + L

g100num [7]

Neutralization reaction means the reactants must be one acid and one alkali, and the product will be H2O and metal salt.

The only one satisfying this will be B

3 0

2 years ago

An ionized helium atom has a mass of 6.6 × 10-27 kg and a speed of 5.3 × 105 m/s. It moves perpendicular to a 0.78-T magnetic fi

arsen [322]

Explanation:

In a magnetic field, the radius of the charged particle is as follows.

r = $\frac{mv}{qB}$

where, m = mass, v = velocity

q = charge, B = magnetic field

Therefore, q will be calculated as follows.

q = $\frac{mv}{rB}$

= $\frac{6.6 \times 10^{-27} \times 5.3 \times 10^{5}}{0.014 m \times 0.78 T}$

= $\frac{34.98 \times 10^{-22}}{0.01092}$

= $3.2 \times 10^{-19} \times \frac{1.0 e}{1.6 \times 10^{-19}C}$

= +2e

Thus, we can conclude that the charge of the ionized atom is +2e.

5 0

2 years ago

Given the equation Ca + H2O --> Ca(OH)2 + H2 how many grams of calcium will react completely with 10.0 grams of water? *

Yuki888 [10]

Answer: 11.0 g of calcium will react with 10.0 grams of water.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $H_2O$

$\text{Number of moles}=\frac{10.0g}{18g/mol}=0.55moles$

The balanced chemical equation is:

$Ca+2H_2O\rightarrow Ca(OH)_2+H_2$

According to stoichiometry :

2 moles of $H_2O$ require = 1 mole of $Ca$

Thus 0.55 moles of $H_2O$ require= $\frac{1}{2}\times 0.55=0.275moles$ of $Ca$

Mass of $Ca=moles\times {\text {Molar mass}}=0.275moles\times 40g/mol=11g$

Thus 11.0 g of calcium will react with 10.0 grams of water.

3 0

2 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

Which type of galaxy forms when two other galaxies crash into each other? They are small galaxies made up of bright, young stars

GenaCL600 [577]

Answer:

elliptical galaxy

Explanation

:The combination of the two galaxies then forms what appears to be an elliptical galaxy as the arms begin to disappear. The merger of gasses creates new stars, and the new shape becomes more elliptical, globular, or sometimes irregular.

4 0

2 years ago