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7 0

Removing seed casings from grains is SEPARATING. a soda bubble bubbling when it is opened is MIXING. a bright copper statue turning green is MIXING. remove salt from seawater is SEPARATING. water decomposing is SEPARATING.

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What is the mass of a silver coin that contains 1.69x10^23 silver atoms? Report your answer in units of grams

Margaret [11]

Answer: 30.27 g

Explanation:

Since the silver is in units of atoms, we have to use Avogadro's number and molar mass.

Avogadro's number: 6.022×10²³ atoms/mol

Molar Mass: 107.87 g/mol

Now that we have everything we need, we can convert atoms to grams.

$1.69*10^{23} atoms*\frac{1mol}{6.022*10^2^3 atoms} *\frac{107.87 g}{1mol} =30.27 g$

3 0

3 years ago

Preparation of Standard Buffer for Calibration of a pH Meter The glass electrode used in commercial pH meters gives an electrica

lana66690 [7]

Answer:

Mass of NaH₂PO₄·H₂O = 8.542 g
Mass of Na₂HPO₄ = 5.410 g

Explanation:

Keeping in mind the equilibrium:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺

We use the Henderson-Hasselbalch equation (H-H):

pH = pka + $log\frac{[A^{-}]}{[HA]}$

For this problem [A⁻] = [HPO₄⁻²] and [HA] = [H₂PO₄⁻]

From literature we know that pka = 7.21, from the problem we know that pH=7.00 and that

[HPO₄⁻²] + [H₂PO₄⁻] = 0.100 M

From this equation we can express [H₂PO₄⁻] in terms of [HPO₄⁻²]:

[H₂PO₄⁻] = 0.100 M - [HPO₄⁻²]

And then replace [H₂PO₄⁻] in the H-H equation, in order to calculate [HPO₄⁻²]:

$7.00=7.21+log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]} \\-0.21=log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\10^{-0.21} =\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\0.616*(0.100M-[HPO4^{-2}])=[HPO4^{-2}]\\0.0616 M = 1.616*[HPO4^{-2}]\\0.03812 M =[HPO4^{-2}]$

With the value of [H₂PO₄⁻], we calculate [HPO₄⁻²]:

[HPO₄⁻²] + 0.0381 M = 0.100 M

[HPO₄⁻²] = 0.0619 M

Finally, using the concentrations, the volume, and the molecular weights; we can calculate the weight of each substance:

Mass of NaH₂PO₄·H₂O = 0.0619 M * 1 L * 138 g/mol = 8.542 g

Mass of Na₂HPO₄ = 0.0381 M * 1 L * 142 g/mol = 5.410 g

8 0

4 years ago

How many grams of KBr are required to make 350. mL of a 0.115 M KBr solution?

Vsevolod [243]

0.115 M means that 0.115 moles of KBr are contained in a volume of 1000 ml, therefore a volume of 350 ml will have (0.115 × 0.35) = 04025 moles
From the formula of molarity moles = molarity × volume in liters
1 mole of KBr is equivalent to 119 g
Therefore, the mass = 0.04025 × 119 g = 4.79 g

6 0

4 years ago

Read 2 more answers

16. Mass = 10g

inessss [21]

Actual volume=Final Volume-initial volume

$\\ \sf\longmapsto 50ml-30ml=20ml$