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3 years ago

6. Why are ionic substances not known as molecules?

Chemistry

2 answers:

Elena L [17]3 years ago

8 0

Answer:

So, an ionic compound actually has two (or even more) different species in a single crystal or solution, which exist independently. The only reason why they are together is because of the total net charge, which must be zero.

Vika [28.1K]3 years ago

5 0

Answer: ionic compounds are not known as molecules because they contain aggregate of positive and negative charges which arise from the transfer of electron from the metallic atom to the non-metallic atom. But in covalent substance, there is a sharing of electron pair between the atoms involved and this makes them to form a molecules

Explanation:

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stiv31 [10]

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2 years ago

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3 years ago

How many moles of gas would occupy 22.4 L at 273K in one arm?

andrew-mc [135]

Answer:

1 mol

Explanation:

Using the general gas law equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the provided information in the question;

V = 22.4L

T = 273K

P = 1 atm

R = 0.0821 Latm/molK

n = ?

Using PV = nRT

n = PV/RT

n = (1 × 22.4) ÷ (0.0821 × 273)

n = 22.4 ÷ 22.4

n = 1mol

4 0

2 years ago

Anhydrous copper sulphate can be used in a test for water. What two things will happen when water is added to anhydrous copper s

Misha Larkins [42]

Answer:

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1 year ago

What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?

Sergeu [11.5K]

Answer:

$\large \boxed{\text{b. 1.5 h}}$

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

$\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt$

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

$\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}$

4 0

3 years ago