All categories

3 years ago

Convert 1.5x10^25 atoms of hydrogen into moles of Hydrogen

Chemistry

2 answers:

777dan777 [17]3 years ago

8 0

Moles = # of atoms / 6.023 x 10^23 (Avogadro’s constant)

1.5 x 10^23 / 6.023 x 10^23 = 51.5 moles of Hydrogen

allsm [11]3 years ago

7 0

Sorry man I need some points but I do t even know the answer

You might be interested in

I need help in this :(

alexandr402 [8]

Answer:

ok follow me twitter dm me for answer The Kid D is my twitter

Explanation:

6 0

3 years ago

Hello! Everyone:) I was just wondering if you can help me with my test. <br><br> Points: 99

AVprozaik [17]

Answer:

By their properties ;)

Explanation:

7 0

2 years ago

The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com

Elanso [62]

Answer:

(a) $Ksp=4.50x10^{-7}$

(b) $Ksp=1.55x10^{-6}$

(d) $Ksp=1.05x10^{-22}$

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) $BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)$

$Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}$

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

$Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}$

(B) $Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)$

$Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}$

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

$Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}$

$Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}$

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

$Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}$

(D) $La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)$

$Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}$

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

$Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}$

Best regards.

7 0

4 years ago

An atom has only 2 electrons in its outer shell will most likely try to gain 6 more electrons through bonding.

jolli1 [7]

False. They usually give away their electrons to form ionic bonds

4 0

3 years ago

I need help with science pls help me

ryzh [129]

Answer:

Explanation:

6. When insulating materials rub against each other, they may become electrically charged .

7. Charging by conduction involves the contact of a charged object to a neutral object.

8. Grounding is the process of removing the excess charge on an object by means of the transfer of electrons between it and another object of substantial size.

9. Grounding is the process of removing the excess charge on an object by means of the transfer of electrons between it and another object of substantial size.

6 0

3 years ago