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What elements on the periidic table are metalloids

butalik [34]

Answer:

Boron (B)

Silicon (Si)

Germanium (Ge)

Arsenic (As)

Antimony (Sb)

Tellurium (Te)

Polonium (Po)

8 0

3 years ago

Which of the following is a balanced chemical equation? A, H2O2 → H2O + O2

LenKa [72]

I believe it's D! Hope you can help with my question!

3 0

3 years ago

If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 Liters, what is the temperature in Kelvin(K)

kvv77 [185]

Answer:

205 K (to 3 significant figures)

Explanation:

Assuming that 4 moles of the gas behaves like an ideal gas and obey the kinetic molecular theory.

Let's apply the ideal gas law, pV= nRT.

Here p denotes the pressure of the gas, V is for volume, n is the number of moles of the gas, R is the universal gas constant and T is the temperature.

Substitute the given information into the equation:

5.6 atm ×12 L= 4 mol ×R ×T

Since pressure is in atm and volume is in L, we can use R= 0.08206 L atm K⁻¹ mol⁻¹.

5.6 atm ×12 L= 4 mol ×0.08206 L atm K⁻¹ mol⁻¹ ×T

T= 67.2 ÷0.32824

T= 204.73 (5 s.f.)

T= 205 K (3 s.f.)

4 0

3 years ago

The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is belie

Deffense [45]

Answer: Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

Explanation:

To calculate the number of moles, we use the equation

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For ozone:

Given mass of ozone = 0.827 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

$\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol$

For nitric oxide:

Given mass of nitric oxide = 0.635 g

Molar mass of nitric oxide = 30.01 g/mol

Putting values in above equation, we get:

$\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol$

For the given chemical equation:

$O_3+NO\rightarrow O_2+NO_2$

By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitric oxide

As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ozone is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles

By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitrogen dioxide

Now, calculating the mass of nitrogen dioxide from equation 1, we get:

Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

$0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g$

Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

8 0

4 years ago

Sample Response: The change in her position (from beginning to end), along with the direction, is considered her displacement. T

kykrilka [37]

The sample response given in the question is right.

To find the answer, we need to know more about the distance and displacement.

<h3>How distance differ from displacement?</h3>

Displacement is the shortest distance between the initial and final points of a body.
It is the change in position with a fixed direction.
Displacement is a vector quantity and can be positive, negative or zero values.
Distance is the length of actual path of the body between initial and final positions.
It's a scalar quantity and it can be positive or zero.
The magnitude of displacement is less than or equal to the distance travelled.

Thus, we can conclude that the given sample response is right.

Learn more about distance here:

brainly.com/question/28124225

#SPJ1

3 0

2 years ago