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3 years ago

Which group of elements is characterized as having little to no reactivity? Halogens

Chemistry

2 answers:

m_a_m_a [10]3 years ago

8 0

Which group of elements is characterized as having little to no reactivity?

C. Alkali Metals

finlep [7]3 years ago

4 0

Answer: Noble gases

Explanation: Noble gases constitutes the extreme right position in the periodic table,

They are have the smallest atomic radius and are the most unreactive elements present in the whole periodic table. There outermost shell of electrons that is valence shell of electrons is completely full thus there is no chance of loss or gain of electrons.

Thus noble gas elements are also known as Inert Gases.

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Ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

Molodets [167]

Answer:

Explanation:

What bro

5 0

2 years ago

Read 2 more answers

Please help me.

kirza4 [7]

Taking into account the definition of density, the density of the sample is 0.5 $\frac{g}{mL}$ .

It is necessary yo know that density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density allows you to measure the amount of mass in a certain volume of a substance.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

$density=\frac{mass}{volume}$

In this case, you know:

mass= 6 g
volume= 12 mL

Then, replacing in the definition of density:

$density=\frac{6 g}{12 mL}$

Solving:

density= 0.5 $\frac{g}{mL}$

Finally, the density of the sample is 0.5 $\frac{g}{mL}$ .

Learn more about density:

7 0

3 years ago

Patient added a 17-g measured dose of polyethylene glycol 3350 (MIRALAX) to 180 mL of water to use as a laxative. If the volume

Sergio [31]

Answer: Apparent density of ethylene glycol 3350 = 17g/15.6mL= 1.09g/mL

Specific gravity of mixture = 1.01

Explanation:

The specific gravity of an object or substance is the ratio of the density of that object or substance to that of a reference substance usually water.

Specific gravity = density of substance/density of water

The density of a substance is the ratio of the mass of that substance to is volume.

Density = Mass/volume

Density of water 1.00 g/mL,

Apparent density of ethylene glycol 3350 = mass/volume

volume of ethylene glycol = volume of mixture - volume of water

volume of ethylene glycol = 195.6 - 180) = 15.6mL

Apparent density of ethylene glycol 3350 = 17g/15.6mL= 1.09g/mL

mass of water = density of water * volume of water

mass of water = 1.00g/mL * 180mL = 180 g

Mass of solution = mass of substance + mass of water = (17 + 180)g = 197g

Density of mixture = mass of mixture / volume of mixture

Density of mixture = 197 g / 195.6 mL = 1.01 g/mL

specific gravity of mixture = density of mixture / density of water

specific gravity of mixture = 1.01g/mL / 1.00g/mL

specific gravity of mixture = 1.01

8 0

4 years ago

Given that a chlorine-oxygen bond has an enthalpy of 243 kJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the

Alecsey [184]

Explanation:

The chemical equation is as follows.

$\frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)$

And, the given enthalpy is as follows.

$\frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g)$ ; $\Delta H$ = 102.5 kJ

Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

$\Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}$

102.5 = $[(\frac{1}{2})x + 498] - [(2)(243)]$

102.5 = $(\frac{1}{2})x + 498 - 486$

102.5 - 12 = $\frac{x}{2}$

x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

$\Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})$

x = $[(\frac{1}{2})181 + (\frac{1}{2})498] - 243$

= 339.5 - 243

= 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

8 0

3 years ago

How much of a 400g sample remains after 4 years if a radioactive isotope has a half-life of 2 years?

timama [110]

Answer:

100 g

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 400 g

Time (t) = 4 years

Half-life (t½) = 2 years

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Time (t) = 4 years

Half-life (t½) = 2 years

Number of half-lives (n) =?

n = t / t½

n = 4 / 2

n = 2

Thus, 2 half-lives has elapsed.

Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:

Original amount (N₀) = 400 g

Number of half-lives (n) = 2

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

N = 1/4 × 400

N = 0.25 × 400

N = 100 g

Thus, the amount of the radioactive isotope remaing is the 100 g.

3 0

3 years ago