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3 years ago

How many moles of solute are present in 2 L of a 25M potassium nitrate (KNO3) solution?

Chemistry

1 answer:

Maksim231197 [3]3 years ago

8 0

Mole= Molarity. Volume(L) = (25)(2)= 50 moles

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How many protons, neutrons, and electrons does 35/17 Cl ^ 1- have?

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Answer:

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2 years ago

Based on their locations in the periodic table, which of these elements would you expect to have the greatest atomic radius?

Llana [10]

Cesium have the greatest atomic radius .

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2 years ago

The ________ ion has eight valence electrons. the ________ ion has eight valence electrons. cr6+ ni6+ fe6+ co6+ mn6+

statuscvo [17]

Answer is: Cr⁶⁺.
₂₄Cr 1s²2s²2p⁶3s²3p⁶3d⁵4s¹.
₂₄Cr⁶⁺ 1s²2s²2p⁶3s²3p⁶. 8 valence electrons.
₂₈Ni 1s²2s²2p⁶3s²3p⁶3d⁸4s².
₂₈Ni⁶⁺ 1s²2s²2p⁶3s²3p⁶3d⁴.
₂₆Fe 1s²2s²2p⁶3s²3p⁶3d⁶4s²
₂₆Fe⁶⁺1s²2s²2p⁶3s²3p⁶3d².
₂₇Co 1s²2s²2p⁶3s²3p⁶3d⁷4s²
₂₇Co⁶⁺ 1s²2s²2p⁶3s²3p⁶3d³.
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7 0

3 years ago

Consider the reaction Mg₂Si(s) + 4 H₂O(ℓ) → 2 Mg(OH)₂(aq) + SiH₄(g). How many grams of silane gas (SiH₄) are formed if 25.0 g of

Neko [114]

Answer:

10.60 grams of silane gas are formed.

Explanation:

From the reaction:

Mg₂Si(s) + 4H₂O(l) → 2Mg(OH)₂(aq) + SiH₄(g)

We know that the limiting reactant is Mg₂Si, so to find the mass of SiH₄ formed we need to calculate the number of moles of Mg₂Si:

$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}}$

Where:

m: is the mass of Mg₂Si = 25.0 g

M: is the molar mass of Mg₂Si = 76.69 g/mol

$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}} = \frac{25.0 g}{76.69 g/mol} = 0.33 moles$

Now, the stoichiometric relation between Mg₂Si and SiH₄ is 1:1 so:

$\eta_{Mg_{2}Si} = \eta_{SiH_{4}} = 0.33 moles$

Finally, the mass of SiH₄ is:

$m_{SiH_{4}} = \eta_{SiH_{4}}*M_{SiH_{4}} = 0.33 moles*32.12 g/mol = 10.60 g$

Therefore, 10.60 grams of silane gas are formed.

I hope it helps you!

3 0

3 years ago

Uranium-236 is _____. created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-

nata0808 [166]

<span>Uranium-236 is intermediate nuclei. created by fusion reactions an unstable isotope of uranium created from four hydrogen atoms used in the H-bomb.

Following is the reaction involved in above process:

</span> $^{235}U$ + $^{1}n$ → $^{236}U$ → $^{144}Ba$ + $^{89}Kr$ + 3 $^{1}n$ <span> + 177 MeV
</span>
Here,
$^{235}U$ = Fission material,
$^{1}n$ = projectile,
$^{236}U$ = intermediate nuclei,
$^{144}Ba$ and $^{89}Kr$ = Fission product

8 0

3 years ago

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