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stiv31 [10]
3 years ago
7

Which of the following elements is only made in laboratories?sbcmtetb

Chemistry
1 answer:
liberstina [14]3 years ago
3 0
Its <span>fermium but the answer isnt on there</span>
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Which compound is a polar molecule and has a linear geometry
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Answer:

Linear molecule is a molecule in which atoms are deployed in a straight line (under 180° angle). Molecules with an linear electron pair geometries have sp hybridization at the central atom. An example of linear electron pair and molecular geometry are carbon dioxide (O=C=O) and beryllium hydride BeH2.

Explanation:

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You put 1 gram of salt into 1 liter of water and stir. The resulting liquid is an example of
Wewaii [24]
Instant salt water.........
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The ΔHvap of a certain compound is 49.09 kJ·mol–1 and its ΔSvap is 53.69 J·mol–1·K–1. What is the boiling point of this compound
rusak2 [61]
Hey there !

<span>Convert Joule to KJ :
</span>
1 j ---------------- 0.001 kj
53.69 j ----------- Kj

Kj = 53.69 * 0.001

=>  0.05369 Kj

T = ΔH / <span>ΔS

T = 49.09 / 0.05369

T = 914.32ºC</span>
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3 years ago
2KClO3 —&gt; 2KCl + 302 this equation represents both a ___ reaction AND a(n) ___ reaction.
andriy [413]

Answer:

option D--Decomposition; oxidation-reduction

6 0
3 years ago
A weak monoprotic acid has molar mass 180 g/mol. When 1.00 g of this acid is dissolved in enough water to obtain a 300 mL soluti
Kaylis [27]

<u>Answer:</u> The value of K_a for the given acid is 3.58\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Initial mass of weak monoprotic acid = 1.00 g

Molar mass of weak monoprotic acid = 180 g/mol

Volume of solution = 300 mL

Putting values in above equation, we get:

\text{Molarity of weak monoprotic acid}=\frac{1.00\times 1000}{180\times 300}\\\\\text{Molarity of weak monoprotic acid}=0.0185M

To calculate the hydrogen ion concentration for given pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

pH = 2.62

Putting values in above equation, we get:

2.62=-\log[H^+]

[H^+]=10^{-2.62}=2.40\times 10^{-3}M=0.0024M

The chemical equation for the dissociation of weak monoprotic acid (HA) follows:

                            HA\rightleftharpoons H^++A^-

<u>Initial:</u>              0.0185

<u>At eqllm:</u>        0.0185-x     x     x

Evaluating the value of 'x'

\Rightarrow x=0.0024

So, equilibrium concentration of HA = (0.0185 - 0.0024) = 0.0161 M

Equilibrium concentration of A^- = x = 0.0024 M

The expression of K_a for above equation follows:

K_a=\frac{[H^+][A^-]}{[HA]}

Putting values in above equation, we get:

K_a=\frac{(0.0024)\times (0.0024)}{0.0161}=3.58\times 10^{-4}

Hence, the value of K_a for the given acid is 3.58\times 10^{-4}

5 0
4 years ago
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