Given atomic weight, density, atomic radius, how do you crystal structure

1. In any nuclear reaction, nuclei are transformed in some way.

Grace [21]

<span>1. In any nuclear reaction, nuclei are transformed in some way.

True

2. Two fissionable substances commonly used in nuclear reactors and nuclear weapons include uranium-235 and plutonium-239.

True

3. Fission reactions can be controlled; thus, they are used to create energy and electricity. But fusion reactions are uncontrollable and require too much energy for economical use.

True

4. Fission reactions occur at such high temperatures in the Sun that the Sun appears to glow.

False

Hope this answers the question. Have a nice day.</span>

4 0

3 years ago

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Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) â†’ 2Al2O3 (s) In

jek_recluse [69]

Answer: The percent yield of the reaction is 74 %

Explanation:

To calculate the number of moles, we use the equation:

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

$\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol$

For oxygen gas:

$\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol$

The chemical equation for the reaction of titanium and chlorine gas follows:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = $\frac{3}{4}\times 0.092=0.069mol$ of oxygen

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

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

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of $Al_2O_3$

So, 0.092 moles of aluminium will produce = $\frac{2}{4}\times 0.092=0.046moles$ of $Al_2O_3$

Now, calculating the mass of aluminium oxide:

$\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g$

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

$\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%$

Hence, the percent yield of the reaction is 74 %

6 0

3 years ago

A box of cleaning supplies weighs 15 N. If the box is lifted a distance of 0.60 m, how much work is done?

Nady [450]

Work Done = force x displacement. So in this case the 15N is the force (because weight is a force) and 0.60m is the displacement. Therefore 15 x 0.6 = 9 Joules of work done (btw, work done can also be referred to as energy transferred)

3 0

4 years ago

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Why does the flame go out on the Bunsen burner when the gas is turned off? (use the collision theory in your answer)

Gre4nikov [31]

Answer:

Explanation:

Well the gas is the fuel for the flame of course. The collision theory comes into play when the gas turns on, chemicals collide with one another. Then reactions occur causing the flame. Then when you take away the fuel, the flame stops because there is no atoms or molecules to come together/collide with one another.

Sorry if its wrong or doesn't make sense... Wish you the best of luck on whatever your doing!

8 0

3 years ago

If I add 2.75 g of CaO to 155 mL of distilled water, what will be the concentration of the solution be?

IRINA_888 [86]

M = m/L
moles of CaO = 2.75g / molar mass
2.75g/ 56.08g = 0.049moles CaO
155mL / 1000 = 0.155 L
M= 0.049moles/0.155L
M=0.316

7 0

2 years ago