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3 years ago

What are standard conditions when working with gases?

Chemistry

1 answer:

zloy xaker [14]3 years ago

4 0

Answer:

STP in chemistry is the abbreviation for Standard Temperature and Pressure. STP most commonly is used when performing calculations on gases, such as gas density. The standard temperature is 273 K (0° Celsius or 32° Fahrenheit) and the standard pressure is 1 atm pressure.

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Explain the difference between wavelength and frequency

Katyanochek1 [597]

Answer:

Wavelength (typically measured in nanometers) is the distance between two points in a wave.Frequency (typically measured in Hertz) is the number of waves in a specific time . Frequency and wavelength have both direct and inverse relationships. The crucial difference between frequency and wavelength is that frequency shows the total number of wave oscillations in a given time. As against wavelength specifies the distance between two specific points of a wave.

Explanation:

Frequency is how often something changes per second be it amplitude of a voltage on a wire or be it the bobbing back and forth of a bobblehead. Frequency is how often something moves up and down in a second. If a bobble head moves forward and backward in one second then it has a bobbling frequency of 1 Hertz (Hz). The unit of frequency is Hertz (Hz) or # of cycles or oscillations per second. A wavelength is measured in distance like meters (m). For photons or light or radiowaves the equation is wavelength=speed of light/frequency.

4 0

3 years ago

Based on the table, which city was in danger of flooding?

dalvyx [7]

Answer:

I think the answer is Toledo

8 0

3 years ago

How many atoms are in 25.00 g of B?

klio [65]

Answer:

There are 1.393 x 10²⁴ atoms in 25.00 g of B.

Explanation:

Hey there!

We are given a value, in grams, that we need to convert to a number of atoms.

We can convert grams to atoms by using Avogadro's Number ( $N_A$ ). This number is equivalent to $6.022 \times 10^{23}$ .

This number can be used to convert any values to:

atoms
molecules
formula units
moles

In order to do this problem, we will need to use dimensional analysis (DA). This process allows us to convert from grams to atoms.

We need to set up our ratios in order to work this out. We can use a periodic table to help us through this next part of the problem.

1. Locating the number of moles of B in the sample

We first need to find the amount of moles of boron (B) there are in the sample.

Checking a periodic table, the atomic mass in atomic mass units (amu) is 10.81 amu.

Atomic mass units can easily be converted to grams and these units can be used interchangeably.

Therefore, for each atom of boron, it weighs 10.81 grams to us. This is equivalent to the mass of one mole of boron.

To find the number of moles, we have two possible ratios we can use:

$\displaystyle \frac{1 \ mole \ B}{10.81 \ grams \ B}$

$\displaystyle \frac{10.81 \ grams \ B}{1 \ mole \ B}$

These ratios mean the same thing, but we need to convert our final unit to moles.

We are given a sample in grams, and when dividing our units, we need to keep moles.

Since the first portion of our expression is in grams, we need to have grams in the bottom of our expression.

$\displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}}$

We can now simplify the expression. Our grams B unit will cancel out, so we are therefore left with moles B remaining.

2. Locating the number of atoms in the sample

Now with our equation, we can convert our number of moles that would be solved if we stopped with the above. However, we need to convert to atoms.

We use Avogadro's number and create a ratio with that of moles.

$\displaystyle \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}$

$\displaystyle \frac{1 \text{mole B}}{6.022 \times 10^{23} \text{atoms}}$

We need to cancel out our moles and end with atoms, so we must have moles in the denominator. Therefore, we use the first ratio.

Using our previous expression, we multiply by this new ratio and solve the expression.

$\displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}} \ \times \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}$