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2 years ago

Explain the advantage of using scientific names,

Chemistry

1 answer:

posledela2 years ago

4 0

Answer:

Clarity and precision - these names are unique with each creature having only one scientific name. Helps avoid confusion created by common names. 3. Universal recognition - scientific names are standardised and accepted universally

Explanation:

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

Write Balance chemical reaction for preparation of chlorine with or without application heat

suter [353]

Answer:

2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)

Explanation:

Chlorine is a diatomic halogen gas known for its greenish-yellow colour. It has a pungent smell and is only moderately soluble in water.

It is a very reactive gas and is never found in free state in nature.

Chlorine can be prepared in the laboratory by oxidation of hydrochloric acid using KMnO4 as follows;

2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)

The set up does not need to be heated.

3 0

2 years ago

Why do scientists share the results of experiments

VladimirAG [237]

this way they can make sure that the experiment is correct.

3 0

3 years ago

Read 2 more answers

Chemical bonding for Rb & nitrogen, Need help!!!

LekaFEV [45]

Covalent bonding!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

5 0

2 years ago

A certain system absorbs 350 joules of heat and has 230 joules of work done on it what is the value of

marysya [2.9K]

According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).

(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt

(10.0 g) / (0.0480 mol) = 208.3 g/mol

So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71

7 0

3 years ago