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3 years ago

In order for reflection to occur the waves must an object

Chemistry

1 answer:

Black_prince [1.1K]3 years ago

5 0

Answer: In order for reflection to occur, the wave must “bounce off” an object

Explanation: For example if your looking in the mirror the waves are bouncing off the mirror to reflect your image

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Describe how hydrogen and oxygen form water.

bekas [8.4K]

Hydrogen and Oxygen by themselves have 1 and 6 valence electrons, respectively. 8 valence electrons is stable, so the atoms form bonds with each other to achieve 8 valence e-.

1 H atom + 1 H atom + 1 O atom = 8 valence e-

7 0

3 years ago

Read 2 more answers

A solution of hydrochloric acid of unknown concentration was titrated with 0.10 M NaOH. If a 100.-mL sample of the HCl solution

madam [21]

Answer: The initial pH of the HCl solution is 3

Explanation:

To calculate the concentration of acid, 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 HCl

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

We are given:

$n_1=1\\M_1=?M\\V_1=100mL\\n_2=1\\M_2=0.10M\\V_2=1mL$

Putting values in above equation, we get:

$1\times M_1\times 100=1\times 0.10\times 1\\\\M_1=\frac{1\times 0.10\times 1}{1\times 100}=10M$

1 mole of HCl produces 1 mole of $H^+$ ions and 1 mole of $Cl^-$ ions

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=0.001M$

Putting values in above equation, we get:

$pH=-\log (0.001)\\\\pH=3$

Hence, the initial pH of the HCl solution is 3

6 0

3 years ago

All isotopes of a particular element have the same atomic number. how then do the isotopes of a particular element differ?

White raven [17]

They have less or more neutrons. Isotopes have same number of protons but different number of neutrons.

4 0

4 years ago

Read 2 more answers

Which of the following does NOT involve a physical change?

jasenka [17]

Answer:mixing decomposing

Explanation:

Melting and grinding changes the physical form and mixing decomposing doesn’t

4 0

3 years ago

The density of a metal is 11.4 g/cm3. How much volume (in cm3) would a sample of 30.5 g have?

julia-pushkina [17]

The Concept:

We are given the density of a sample of the metal = 11.4 grams / cm³

and we need to find the volume occupied by a sample of 30.5 grams

For this solution, we will use dimensional analysis

from the given information, we can also say that the density of the metal is:

1 cm³ / 11.4 grams

If we multiply this value by 30.5 grams, the 'grams' in the numerator and the denominator will cancel out and we will be left with the volume occupied by 30.5 grams of the metal

Solving for the volume:

$\frac{ 1 cm^{3} }{11.4 grams}$ X 30.5 grams = (30.5 / 11.4) cm³

Volume of 30.5 grams of the sample = 2.68 cm³

6 0

3 years ago