Explain how mass and energy can be conserved even during a chemical change.

(Please answer ASAP) Describe in your own words A.) what does Inert mean? B.) why are noble gases Inert (include an explanation

Alex_Xolod [135]

Inert means unreactive in nature or un reactive in there normal or natural state
and noble gas has there outermost shell electron filled up they complete there duplet or oclet valency state that why there inert in there natural state

4 0

3 years ago

What is a term called gradient in Earth Science?

goblinko [34]

The gradient is the slope of a linear equation, represented in the simplest form as y = mx + b. In Earth Science, the gradient is usually used to measure how steep certain changes in elevation are.

In order to find the gradient in a topographical setting, one must know two things: the elevation of two points and the distance between the two points. Once these values are known, the gradient can be found by dividing the change in field value, or the change in elevation, by the distance. The higher the gradient value is, the steeper the slope is.

4 0

3 years ago

(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW

Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = 2.3x10⁻³M InF₃. Thus:

[In⁺³] = 2.3x10⁻³M InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M In⁺³

[F⁻] = 2.3x10⁻³M InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M In⁺³] × [7.0x10⁻³M F⁻]³ = 7.9x10⁻¹⁰

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, solubility is 6.31x10⁻⁶M

3 0

3 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago

Eeeeeeeeeeeeeeeeeeeeeeeeeee

Varvara68 [4.7K]

Answer:

dolo HAHAHAAHAHHAHAHAHA

6 0

2 years ago