All categories

3 years ago

How many grams of sucrose must be dissolved in 250 ml to prepare the 0.15 mol / L solution?

Chemistry

1 answer:

Harlamova29_29 [7]3 years ago

6 0

Answer:

How could you prepare 250 mL of 0.20M NaCl using only a solution of 1.0M NaCl and water? To prepare a 250 mL 0.20 M NaCl solution, I would place 50 mL of the 1.0 stock solution to a 250 mL volumetric flask and then add 200 mL of water. A bottle of the antiseptic hydrogen peroxide (H2O2) is labeled 3.0% (v/v).

You might be interested in

Approximately how many elements are there that combine chemically in a great number of ways to produce compounds? A. 25 B. 50 C

Arturiano [62]

100 as there’s approximately 100 discovered elements

8 0

3 years ago

Read 2 more answers

A giant mass of air that carries warm weather to regions of the earth?

Scilla [17]

Continental air masses are characterized by dry air near the surface while maritime air masses are moist .Polar air masses are characterized by cold air near the surface while tropical air masses are warm or hot. Arctic air masses are extremely cold.
:::::::)

6 0

3 years ago

Which of the following would have the largest pKa?

garri49 [273]

Answer:

CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

Explanation:

To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:

pKa of CH3CH2NH3+ = CH3CH2NH2; C6H5NH3+ = C6H5NH2

Also, Kw / Kb = Ka

Thus:

pKa of CH3CH2NH3+/CH3CH2NH2 is:

Kw / kb = Ka = 1.79x10⁻¹¹

-log Ka = pKa

pKa = 10.75

pKa of C6H5NH3+/ C6H5NH2 is:

Kw / kb = Ka = 2.5x10⁻⁵

-log Ka = pKa

pKa = 4.6

That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

5 0

3 years ago

Describe the strength of the Coriolis effect as you move from the equator to the poles of the earth

Alekssandra [29.7K]

The Coriolis Effect describes the turn of the wind to the right in the Northern Hemisphere caused by earth's rotation.

3 0

3 years ago

A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________

gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

Average velocities of gases can be expressed as root-mean-square averages. (V rms)

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

R = gas constant, T = temperature, Mm = molar mass of the gas particles

From the question

R = 8,314 J / mol K

T = temperature

Mm = molar mass, kg / mol

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

$\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s$

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

M₁ = molar mass sulfur dioxide = 64

M₂ = molar mass nitrogen triodide = 395

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5$

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0

2 years ago