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babunello [35]
2 years ago
15

HELP ME PLEASE. no links.

Chemistry
1 answer:
tatiyna2 years ago
4 0

Answer:

Experimental group

Explanation:

i hopes this helps let me know if im wrong:)

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How many micrograms (ug) are in 3.4 x 10^-5 ounces (oz)
marysya [2.9K]

Answer:

964ug

Explanation:

The problem here involves converting from one unit to another.

 We are to convert from ounces to micrograms.

                                    1ug  = 1 x 10⁻⁶g

                                    1oz  = 28.35g

       

So we first convert to grams from oz then take to ug:

 Solving:

                    1oz  = 28.35g

             3.4 x 10⁻⁵oz  will then give  3.4 x 10⁻⁵ x 28.35 = 9.64  x 10⁻⁴g

So;

                    1 x 10⁻⁶g    = 1ug

          9.64  x 10⁻⁴g will give \frac{9.64 x 10^{-4} }{1 x 10^{-6} }      = 9.64 x 10²ug or 964ug

8 0
2 years ago
How does the density of a gas depend on the molar mass of the gas?
AlladinOne [14]

Answer:

The density of the ideal gas is directly proportional to its molar mass.

Explanation:

Density is a scalar quantity that is denoted by the symbol ρ (rho). It is defined as the ratio of the mass (m) of the given sample and the total volume (V) of the sample.

\rho = \frac{m}{V}                          ......equation (1)

According to the ideal gas law for ideal gas:

PV = nRT                                       ......equation (2)

Here, V is the volume of gas, P is the pressure of gas, T is the absolute temperature, R is Gas constant and n is the number of moles of gas

As we know,

The number of moles: n = \frac{m}{M}

where m is the given mass of gas and M is the molar mass of the gas

So equation (2) can be written as:

PV = \frac{m}{M}RT

⇒ PM= \frac{m}{V} RT

⇒ \frac{PM}{RT}= \frac{m}{V}             ......equation (3)

Now from equation (1) and (3), we get

\frac{PM}{RT}= \frac{m}{V} = \rho  

⇒ Density of an ideal gas: \rho = \frac{PM}{RT}  

⇒ <em>Density of an ideal gas: ρ ∝ molar mass of gas: M</em>

<u>Therefore, the density of the ideal gas is directly proportional to its molar mass. </u>

6 0
3 years ago
A wave has a frequency of 3,5MHz and a wavelength of 85,5 m. What is the wave velocity through
Yuki888 [10]

Answer:

velocity = 29925×10⁶ m/s

Explanation:

Given data:

Frequency = 35 MHz

Wavelength = 855 m

Velocity = ?

Solution:

MHz to Hz

35×10⁶ Hz

Formula:

<em>velocity = Wavelength × Frequency</em>

velocity = 855 m × 35×10⁶ Hz

velocity = 29925×10⁶ m/s

Hz = s⁻¹

5 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
Which of these changes do not involve forming or breaking chemical bonds?
gregori [183]
D
Boiling and melting are both physical changes so do not involve the formation or breakage of chemical bonds.
5 0
3 years ago
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