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3 years ago

How many H2O molecules are contained in 0.0016 moles of the hydrate Na2B4O7.10H2O?

2 answers:

8 0

Maybe this example could help you to understand this problem.

https://image.slidesharecdn.com/121howmanyatoms-091201144624-phpapp02/95/12-1-how-many-atoms-17-728....

MaRussiya [10]3 years ago

5 0

https://image.slidesharecdn.com/121howmanyatoms-091201144624-phpapp02/95/12-1-how-many-atoms-17-728....

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The first two columns in the periodic table react with? A) Gases B) Metals C) Water

timofeeve [1]

It should be A because the periodic table starts with Hydrogen in this one is a gas

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3 years ago

Read 2 more answers

Which is the correct electron dot diagram for krypton (Kr)?

SCORPION-xisa [38]

Krypton is in group 18, and so it has 8 valence electrons. B is the answer

4 0

4 years ago

Calculate the solubility (in g/L) of CaSO 4 ( s ) CaSO4(s) in 0.400 M Na 2 SO 4 ( aq ) at 25 ° C 0.400 M Na2SO4(aq) at 25°C. The

Allushta [10]

Explanation:

Ionization equation for $CaSO_{4}$ is as follows.

$CaSO_{4} \rightarrow Ca^{2+} + SO^{2-}_{4}$

s s s

Now, the expression for the solubility product is as follows.

$K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

= $s \times s$

= $s^{2}$

As the concentration of $Na_{2}SO_{4}$ is given as 0.4 M.

So, $[Na_{2}SO_{4}] = [SO^{2-}_{4}]$ = 0.4 M

Putting the given values as follows.

$K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

$4.93 \times 10^{-5} = [Ca^{2+}] \times 0.4$

$[Ca^{2+}]$ = 12.325 \times 10^{-5}[/tex]

Hence, the solubility of $CaSO_{4}$ in $Na_{2}SO_{4}$ is $12.325 \times 10^{-5}$ .

Therefore, solubility of $CaSO_{4}$ in g/ml as follows.

$12.325 \times 10^{-5} \times 136 g/mol$

= 0.0167 g/L

Thus, we can conclude that solubility of $CaSO_{4}$ is 0.0167 g/L.

4 0

3 years ago

Given: glow is a find the indicated measure.

Alisiya [41]

Answer:

is there supposed to be some type of image or ...

Explanation:

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3 years ago

calculate atomic weight of silver which has two isotopes with the following properties : silver-107 (106.91 amu, 51.84% natural

Julli [10]

Answer: The atomic weight of silver is 107.87 amu

Explanation:

Mass of isotope silver-107 = 106.91 amu

% abundance of isotope silver -107 = 51.84% = $\frac{51.84}{100}=0.5184$

Mass of isotope silver-109 = 108.90 amu

% abundance of isotope silver-109 = 48.16% = $\frac{48.16}{100}=0.4816$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(106.91\times 0.5184)+(108.90\times 0.4816)]$

$A=107.87amu$

Thus atomic weight of silver is 107.87 amu

8 0

4 years ago