Graph the circle which is centered at (4,4)(4,4)left parenthesis, 4, comma, 4, right parenthesis and has a radius of 3 units.

The energy levels of the positive ions that have just a single electron, like He , Li2 , or Be3 , etc., follow the formula Bohr

Sidana [21]

The answer is false. Hope that helps

3 0

3 years ago

What risks does nuclear waste pose to Earth or the sun or outer space?

sesenic [268]

Answer:

It could pollute other planets. Nuclear waste might be attracted by the gravitational pull of the sun and disintegrate there instead. The waste storage spacecraft could crash into a planet, moon, or another astronomical body. It could be dangerous for other life forms out there.

Explanation:

6 0

3 years ago

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) What is the max

jolli1 [7]

Answer: The mass of aluminium chloride that can be formed are 46.3 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 Aluminium:

Given mass of aluminium = 32 g

Molar mass of aluminium = 26.98 g/mol

Putting values in above equation, we get:

$\text{Moles of aluminium}=\frac{32g}{26.98g/mol}=1.186mol$

For Chlorine:

Given mass of chlorine = 37 g

Molar mass of chlorine = 71 g/mol

Putting values in above equation, we get:

$\text{Moles of chlorine gas}=\frac{37g}{71g/mol}=0.521mol$

For the given chemical equation:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)$

By Stoichiometry of the reaction:

3 moles of chlorine gas is reacting with 2 moles of aluminium.

So, 0.521 moles of chlorine gas will react with = $\frac{2}{3}\times 0.521=0.347moles$ of aluminium.

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

So, chlorine gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of chlorine gas is producing 2 moles of aluminium chloride

So, 0.521 moles of chlorine gas will react with = $\frac{2}{3}\times 0.521=0.347moles$ of aluminium chloride.

Now, calculating the mass of aluminium chloride by using equation 1, we get:

Moles of aluminium chloride = 0.347 moles

Molar mass of aluminium chloride = 133.34 g/mol

Putting all the values in equation 1, we get:

$0.347mol=\frac{\text{Mass of aluminium chloride}}{133.34g/mol}\\\\\text{Mass of aluminium chloride}=46.3g$

Hence, the mass of aluminium chloride that can be formed are 46.3 g

7 0

4 years ago

Which statement best describes the effect of beta particles on body tissue?

mash [69]

I think the answer is C

3 0

3 years ago

Read 2 more answers

What is the empirical formula of a compound that contains 49.4% K, 20.3% S, and 30.3% by mass? A) KSO3 B) K2SO3 C) KSO2 D) KSO E

yulyashka [42]

Answer: $K_2SO_3$ .

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of K = 49.4 g

Mass of S = 20.3 g

Mass of O = 30.3 g

Step 1 : convert given masses into moles.

Moles of K= $\frac{\text{ given mass of K}}{\text{ molar mass of K}}= \frac{49.4g}{40g/mole}=1.23moles$

Moles of S= \frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{20.3g}{32g/mole}=0.63moles[/tex]

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{30.3g}{16g/mole}=1.89moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For K = $\frac{1.23}{0.63}=2$

For S = $\frac{0.63}{0.63}=1$

For O = $\frac{1.89}{0.63}=3$

The ratio of K: S:O = 2: 1: 3

Hence the empirical formula is $K_2SO_3$ .

7 0

4 years ago