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Romashka-Z-Leto [24]
3 years ago
7

This element has an atomic number smaller than magnesium and 3 electrons in its valence shell

Chemistry
1 answer:
OverLord2011 [107]3 years ago
6 0

Boron (B) is your answer.

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Air resistance depends on two important factors - the speed of the object and it's surface area. Increasing the surface area of an object decreases it's speed
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This is a material that allows heat/electricity to transfer.
Anvisha [2.4K]
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5 0
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QUESTION 3 (a) Ammonium sulphate, (NH),50, is a soluble salt and it is used in agriculture as fertiliser. 5 g of ammonium sulpha
nataly862011 [7]

Answer:

The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)

The number of moles = 5 g / 132.14 g/mol = 0.038 mol

The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23

the number of positive ions present in the ammonium sulphate solution:

2 positive ions for every 1 molecule of (NH₄)₂SO₄

so 2 x 2.29x10^23 = 4.58x10^23

the number of negative ions present in the ammonium sulphate solution

1 negative ion for every 1 molecule of (NH₄)₂SO₄

so 1 x 2.29x10^23 = 2.29x10^23

the total number of ions present in the ammonium sulphate solution​

4.58x10^23 + 2.29x10^23 = 6.87x10^23

4 0
3 years ago
You wish to construct a buffer of pH=7.0. Which of the following weak acids (w/ corresponding conjugate base) would you select?
Alex17521 [72]

Answer:

C.) HOCl Ka=3.5x10^-8

Explanation:

In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below

we Know that

pKa= -log(Ka)

therefore

A) pKa of  HClO2 = -log(1.2 x 10^-2)

=1.9208

B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644

C)  pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45

D) pKa of HCN = -log(4 x 1 0^-10)=  9.3979

If we consider the  Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution

The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.

So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.

Hence, HOCl will be chosen for buffer construction.

3 0
3 years ago
Consider the following types of electromagnetic radiation:
liq [111]

Explanation:

Electromagnetic wave              Wavelength

(1) Microwave  =   1 m to 1 mm = 10^9 nm to 10^6 nm

(2) Ultraviolet  =    10 nm to 400 nm

(3) Radio waves  =   1 mm to 100 km = 10^6 nm to 10^{14}nm

(4) Infrared  =    700 nm to 1 mm

(5) X-ray  =   0.01 nm to 10 nm

(6) Visible =   400 nm t0 700 nm

a) In order of increasing wavelength:

: 5 < 2 < 6 < 4 < 1 < 3

b) Frequency of the electromagnetic wave given as:

\nu=\frac{c}{\lambda }

\nu = frequency

\lambda = Wavelength

c = speed of light

\nu \propto \frac{1}{\lambda }

So, the increasing order of frequency:

: 3 < 1 < 4 < 6 < 2 < 5

c) Energy(E) of the electromagnetic wave is given by Planck's equation :

E=h\nu

E\propto \nu

So, the increasing order of energy:

: 3 < 1 < 4 < 6 < 2 < 5

6 0
3 years ago
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