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3 years ago

Which 3 elements had complete out shells? (you will probably need to use the periodic table)

Chemistry

1 answer:

wlad13 [49]3 years ago

3 0

The inert gases like neon,argon ,krypton have complete shells

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2.(04.01 LC)

Mila [183]

Answer:

2KClO3 —> 2KCl + 3O2

The coefficients are 2, 2, 3

Explanation:

From the question given above, we obtained the following equation:

KClO3 —> 2KCl + 3O2

The above equation can be balance as follow:

There are 2 atoms of K on the right side and 1 atom on the left side. It can be balance by putting 2 in front of KClO3 as shown below:

2KClO3 —> 2KCl + 3O2

Now, the equation is balanced.

Thus, the coefficients are 2, 2, 3

8 0

3 years ago

What is the ionic charge for the calcium ion in CaCO3?

Lunna [17]

<span>The ionic charge of Calcium (Ca) in calcium carbonate (CaCO3) is 2+. CaCO3 has a neutral ionic charge sin CO3 has a 2- charge.</span>

3 0

3 years ago

Convert 100mL to L. Show your work to receive full credit.

Ilya [14]

Answer:

100 ml = 0.1 L

I divided 100 by 1000 because 1000ml = 1L

3 0

3 years ago

Predict: Will a 1500 kg car moving at the same speed as the tennis ball have

sergeinik [125]

The car will have more momentum because it has a greater mass. Momentum= mass x velocity therefore if both the objects have the same velocity the object with a greater mass will have more momentum

4 0

3 years ago

A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

6 0

3 years ago