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4 years ago

What was the normal perception of a scientist up until the seventeenth century?

1 answer:

7 0

working in a factory to develop new technology wearing a white lab coat working in a laboratory working in the woods to study nature collaborating with groups of peers to discuss ideas

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What is meant by atomic radius

Kaylis [27]

Answer:

The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons. ... The value of the radius may depend on the atom's state and context.

Explanation:

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4 years ago

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Calculate the maximum numbers of moles and grams of iodic acid (HIO₃) that can form when 635 g of iodine trichloride reacts with

poizon [28]

What is Chemical Reaction?

A chemical reaction occurs when a certain group of molecules converts into another form without affecting their nuclei; only the transfer or sharing of electrons and the building and breaking of bonds occur.

Main Content

Known :

$\mathrm{ICl}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ICl}+\mathrm{HIO}_{3}+\mathrm{HCl}$

Mass of $ICI_3 = 635g$

Mass of $H_2O = 118.5g$

Calculations :

First, balance the given chemical equation by place 2,3 and 5 as the coefficients of $ICI_3, H_2O and HCl,\\$ Respectively

$2 \mathrm{ICl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ICl}+\mathrm{HIO}_{3}+5 \mathrm{HCl}$

We multiply the given mass of $ICl_3$ by the reciprocal of its molar mass to get the number of moles. The molar mass of $ICl_3 is 233.26g/mol$

$\text { Moles of } \mathrm{ICl}_{3}=635 \mathrm{~g} \mathrm{ICl}_{3} \times \frac{1 \mathrm{~mol} \mathrm{ICl}_{3}}{233.26 \mathrm{~g} \mathrm{ICl}_{3}}=2.7223 \mathrm{~mol} \mathrm{ICl}_{3}$

Them, we multiply the ratio between $ICl_3 and HIO_3$ . based on the chemical equation, the molar ratio 1 mol $HIO_3/2 mol ICl_3$ .

$\text { Moles of } \mathrm{HIO}_{3} \text { formed }=2.7223 \mathrm{~mol} \mathrm{ICl}_{3} \times \frac{1 \mathrm{~mol} \mathrm{HIO}_{3}}{2 \mathrm{~mol} \mathrm{ICl}_{3}}=1.3612 \mathrm{~mol} \mathrm{HIO}_{3}$

For water, we again multiply the given mass $H_2O$ by the reciprocal of its molar mass to get the number if moles. The molar mass of $H_2O$ is 18.02g/mol

$\text { Moles of } \mathrm{H}_{2} \mathrm{O}=118.5 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} \times \frac{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{18.02 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}}=6.5760 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}$

Then, we multiply the molar ratio between H2O and HIOs. Based on the chemical equation, the molar ratio is 1 mol $HIO_3/3 Mole H_2O$

$\text { Moles of } \mathrm{HIO}_{3} \text { formed }=6.5760 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O} \times \frac{1 \mathrm{~mol} \mathrm{HIO}_{3}}{3 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}=2.1920 \mathrm{~mol} \mathrm{HIO}_{3}$

We can see that the limiting reactant is $ICL_3$ since the given mass of $ICl_3$ forms lesser product than water does. Thus , the maximum number of moles of $HIO_3$ formed is 1.36 mol $HIO_3$ . We now multiply the molar mas of $HIO_3$ to the calculated number of moles. The molar mass of $HIO_3$ is 175.91 g/mol.

Mass of $HIO_3$ formed (Max) $=1.3612 \mathrm{~mol} \mathrm{HIO}_{3} \times \frac{175.91 \mathrm{~g} \mathrm{HIO}_{3}}{1 \mathrm{~mol} \mathrm{HIO}}=239 \mathrm{~g} \mathrm{HIO}_{3}$

We multiply the number of moles of $ICl_3$ by the molar ratio between $ICl_3$ and $H_2O$ which is $3 mol H_2O mol ICl_3$ we get the number of moles of $H_2O$ reacted. Then, we multiply the molar mass of water.

Mass of $H_2O$ reacted $=2.7223 \mathrm{~mol} \mathrm{ICl} 3 \times \frac{3 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{2 \mathrm{~mol} \mathrm{ICl}_{3}} \times \frac{18.02 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}$

Mass of $H_2O$ reacted = 73.6g $H_2O$

We subtract the mass of $H_2O$ reacted from the given mass of $H_2O$ .

Mass of $H_2O$ = 118.5 - 73.6g = $44.9g H_2O$

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2 years ago

Identify whether longhand notation or noble-gas notation was used in each case below. Potassium (K): 1s22s22p63s23p64s1

Vaselesa [24]

<u>Answer:</u> The given electronic configuration is long hand notation.

<u>Explanation:</u>

Long-hand notation of representing electronic configuration is defined as the arrangement of total number of electrons that are present in an element.

Noble-gas notation of representing electronic configuration is defined as the arrangement of valence electrons in the element. The core electrons are represented as the previous noble gas of the element that is considered.

The given electronic configuration of potassium (K): $1s^22s^22p^63s^23p^64s^1$

The above configuration has all the electrons that are contained in the nucleus of an element. Thus, this configuration is a long-hand notation.

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3 years ago

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To construct the galvanic cell illustrated above, the salt bridge was prepared by soaking a piece of cotton in 5.0MNaNO3(aq) bef

dalvyx [7]

Answer:

The cell reaction reaches equilibrium quickly and the cell emf becomes zero.

Explanation:

The purpose of a salt bridge is not to move electrons from the electrolyte, its main function is to maintain charge balance because the electrons are moving from one-half cell to the other.

A solution of a salt that dissociates easily is normally used. Water is ineffective at functioning as a salt bridge. Hence the effect stated in the answer.

4 0

4 years ago

It is common for students to overshoot the endpoint, meaning they add too much NaOH(aq) from the buret, which causes the solutio

umka2103 [35]

Answer: the percentage of acetic acid will be low.

Explanation: The major aim during titration of acids and bases is to determine the endpoint , that is exact point where the acid in the beaker changes colour, (in this case, pink )with an additional drop from the burette containing the base, since it is usually difficult to mark the equivalence point that tells us when all the substrate in the beaker has been neutralized completely with the buretted substance.

Overshooting the end point is an error which can occur when the person involved in the the titration accidently goes beyond this endpoint by adding too much of the substance(base) from the burette into the beaker missing the exact endpoint.

This implies that the person has added too much of the burreted liquid, ie the base than required , making the acid in the beaker to continue to react resulting to a lower concentration of the acid (acetic acid) with excess base.(NaOH)

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3 years ago