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3 years ago

2.14181 x 10^7 micro gram

Chemistry

1 answer:

Finger [1]3 years ago

6 0

Answer:

21418100

Explanation:

2.14181 are remove point then give 14181/100000 then 10^7= 10000000 and solve it 14181/100000*10000000

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Consider the electric field lines for two point charges separated by a small distance. what are the signs of q1 and q2?

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Assume positive charges

directions outwards from source points like field lines of two repelling magnets

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4 years ago

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

$Number \, of \, moles, n = \frac{Mass}{Molar \ mass} = \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl$

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

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4 years ago

A tool that can be measure of force

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A dynamometer is used in measuring force!

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4 years ago

Read 2 more answers

when the volume of a gas is changed from 250 mL to 425 mL, the temperture will change from 137 degrees celsius to _____ degrees

FrozenT [24]

Charles law gives the relationship between temperature and volume of gas.
It states that at a constant pressure, volume of gas is directly proportional to the temperature for a fixed amount of gas.
V/T = constant
where V- volume, T- temperature
V1/T1 = V2/T2
Where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
temperature should be in Kelvin for calculation
temperature in K = temperature in Celcius + 273
T1 = 137 °C + 273 = 410 K
substituting values in the equation
250 mL / 410 K = 425 mL /T2
T2 = 697 K
temperature in Celcius = 697 K - 273
temperature = 424 °C

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3 years ago

A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research techni

yarga [219]

Answer:

The pKa of X-281 is 3.73.

Explanation:

X-281 is a monoprotic weak acid so that it will not dissociate completely in solution (weak acid) and will only produce 1 mol of protons per mol compound (monoprotic acid).

The dissociation equation could be written as follows:

X-281-H ⇆ X-281 + H⁺

Note the equilibrium arrows indicating that not all X-281-H dissociates at equilibrium.

Initially, the concentration of X-281-H is 0.089 M. At equilibrium, the concentration of the dissociation products, X-281 and H⁺, is unknown but both must be the same, since the drug is a monoprotic acid. We can call "X" to the concentration of the products. The concentration of X-281-H is the initial concentration minus the concentration of the products: 0.089 M - X. Then, at equilibrium, these are the concentrations of each species present:

X-281-H ⇆ X-281 + H⁺

(0.089 M-X) X X

We also know that the pH is 2.40. Then:

pH = -log[H⁺] = 2.40

where [H⁺] is the molar concentration of the protons or "X". Then:

-log X = 2.40

log X = -2.40

X = 10^(-2.40) = 3.98 x 10⁻³ M

The concentration of each species present in the equilibrium is then:

[H⁺] = 3.98 x 10⁻³ M

[X-281] = 3.98 x 10⁻³ M

[X-281-H] = 0.089 M - 3.98 x 10⁻³M = 0.085 M

At equlibrium, the acidity constant Ka is:

Ka = [X-281] * [H⁺] / [X-281-H]

Ka = (3.98 x 10⁻³ M * 3.98 x 10⁻³ M ) / 0.085 M = 1.86 x 10⁻⁴

Then the pKa is:

pKa = -log Ka = -log (1.86 x 10⁻⁴) = <u>3.73</u>

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4 years ago