Câu 47: Cách pha chế dung dịch từ hoá chất tinh khiết, phát biểu sai:

Question 25 of 33

Dafna11 [192]

Answer:

A) It tends to increase from top to bottom of a group

6 0

3 years ago

Which chemical property deals with the ability of a substance being able to

Fantom [35]

$\large\underline{\bf\red{Solution:-}}$

The chemical property deals with the ability of a substance being able to burn is called combustibility [ option b ] .

More to know :-

1) Combustion :-

Combustion is a chemical process in which a substance reacts rapidly with oxygen and gives out heat.

The fuel can be any one of the 3 states of matter that is - solid , liquid & gases .

Examples of combustible substances :-

Solid - Coal

Liquid - Petrol , Disel , etc.

Gas - Hydrogen , Oxygen , etc .

5 0

3 years ago

Read 2 more answers

Which process occurs in an operating voltaic cell ?

Alex

Please identify and specify the question please!

8 0

3 years ago

If He gas has an average kinetic energy of 4310 J/mol under certain conditions, what is the root mean square speed of O2 gas mol

Free_Kalibri [48]

Answer:

The root mean square speed of O2 gas molecules is

519.01 m/s

Explanation:

The root mean square velocity :

$v_{rms}=\sqrt{\frac{3RT}{M}}$

$K.E_{avg}=\frac{3}{2}RT$

$K.E =\frac{1}{2}mv_{rms}^{2}$

Molar mass , M

For He = 4 g/mol

For O2 = 2 x 16 = 32 g/mol

O2 = 32/1000 = 0.032 Kg/mol

First calculate the temperature at which the K.E of He is 4310J/mol

K.E of He =

$K.E_{avg}=\frac{3}{2}RT$

$T=\frac{2(K.E)}{3(R)}$

K.E of He = 4310 J/mol

$T=\frac{2(4310J/mol)}{3(8.314J/Kmol)}$

$T=345.60K$

Now , Use Vrms to calculate the velocity of O2

$v_{rms}=\sqrt{\frac{3(8.314J/Kmol)(345.60K)}{0.032Kg/mol}}$

$v_{rms}=\sqrt{\frac{8619.9552}{0.032}}$

$v_{rms}=\sqrt{26935.001}$

$v_{rms}=519.01m/s$

6 0

3 years ago

A 50.00 mL sample of groundwater is titrated with 0.0300 M EDTA . Of 12.40 mL of EDTA is required to titrate the 50.00 mL sample

Rzqust [24]

Answer:

7.44x10⁻³ mol/L and 744 ppm

Explanation:

Let's assume that the hardness of the water is totally from Ca⁺² ions only(the hardness is the measure of Ca⁺² and Mg⁺² ions). The titration with EDTA will form a complex. The EDTA is always in 1:1 proportion, so the number of moles of it will be the number of moles of Ca⁺², which will be the number of moles of CaCO₃.

n = 0.0124 L * 0.0300 mol/L

n = 3.72x10⁻⁴ mol

The molarity is the number of moles divided by the volume (0.05 L)

M = 3.72x10⁻⁴/0.05

M = 7.44x10⁻³ mol/L

1 part per million = 1 mg/L. The molar mass of the CaCO₃ is 100 g/mol, so the mass of it is:

m = 3.72x10⁻⁴ mol * 100 g/mol

m = 0.0372 g = 37.2 mg

Then, the ppm:

37.2/0.05 = 744 ppm

5 0

3 years ago