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How many ug of nickel (Ni) are required to make 25.00 nanoliters of a 1.25 mol/L solution? Be sure to report your answer to the

devlian [24]

volume of Ni = 25 nL = 25 x 10⁻⁹ L

mol Ni = 25 x 10⁻⁹ L x 1.25 mol/L = 3.125 x 10⁻⁸

mass = mol x Ar Ni

mass = 3.125 x 10⁻⁸ x 59 g/mol

mass = 1.84 x 10⁻⁶ g = 1.84 μg

4 0

2 years ago

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How much did bottled water grow between 2015 and 2016?

Solnce55 [7]

Answer:

The International Bottled Water Association (IBWA), Alexandria, Va., and Beverage Marketing Corporation (BMC), New York, recently released 2015 bottled water statistics showing that Americans' consumption of bottled water increased by 7.9 percent and bottled water sales were up 8.9 percent from the previous year.

Explanation:

7 0

3 years ago

True or false: the Molecules in a solid have MORE thermal energy than the molecules in a gas?

katrin2010 [14]

That's false. Because, the molecules in a solid are compacted too tightly together that they don't have all that extra space that a gas would. Molecules in a gas would have more thermal energy because they have all that space to move around in. The more they move around, the hotter it gets. The less that they move around, and the more that they put and packed together, the colder it's going to be.

Hope this helped you!! (:

3 0

3 years ago

Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid

jasenka [17]

Answer: a) The $K_a$ of acetic acid at $25^0C$ is $1.82\times 10^{-5}$

b) The percent dissociation for the solution is $4.27\times 10^{-3}$

Explanation:

$CH_3COOH\rightarrow CH_3COO^-H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.10 M and $\alpha$ = ?

Also $pH=-log[H^+]$

$2.87=-log[H^+]$

$[H^+]=1.35\times 10^{-3}M$

$[CH_3COO^-]=1.35\times 10^{-3}M$

$[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M$

Putting in the values we get:

$K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}$

$K_a=1.82\times 10^{-5}$

b) $\alpha=\sqrt\frac{K_a}{c}$

$\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}$

$\alpha=4.27\times 10^{-5}$

$\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}$

5 0

3 years ago

Silver has two naturally isotopes and has an atomic mass of 107.868 amu. One isotope is Ag-109 isotope (108.905 amu) and has a n

Triss [41]

Answer: 106.905

Explanation: If there are only 2 isotopes, and 1 of them is 48.16%, the second must, by default, be (100 - 48.16%) = 51.84% The final, averaged, atomic mass is 107.868. This is made up of each isotope's atomic mass times the percentage of that isotope in the total sample. The weighted value of the known isotope (109) plus that of the unknown must come to the observed value of 107.868 amu. (107.868 - 52.45 = 55.42). Divide that by the % for that isotope (55.42/0.5184) = 106.90 amu for the second isotope.

Atomic Mass % of Sample Weighted Value

108.905 48.16% 52.45

X 51.84% 55.42

107.87

X = (55.42/0.5184) = 106.90 amu

5 0

3 years ago