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3 years ago

Which periodic table trend is represented by the blue arrows?

Chemistry

1 answer:

blsea [12.9K]3 years ago

3 0

Answer:

Periodic trends are specific patterns that are present in the periodic table that illustrate different aspects of a certain element, including its size and its electronic properties. Major periodic trends include: electronegativity, ionization energy, electron affinity, atomic radius, melting point, and metallic character. Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool to quickly predict an element's properties. These trends exist because of the similar atomic structure of the elements within their respective group families or periods, and because of the periodic nature of the elements.

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How many moles of water will be generated during the combustion of 0.38 moles of methyl alcohol (CH3OH)? 2CH3OH + 3O2 2CO2 + 4H2

ELEN [110]

You already got the balanced equation. So the ratio of mole number is the ratio of the coefficient. Then the ratio of methyl alcohol and water is 2:4=1:2. The water generated is 0.38*2=0.76 mol.

6 0

3 years ago

A wooden object from the site of an ancient temple has a carbon-14 activity of 10 counts/min compared with a reference piece of

Arada [10]

Answer:

Age of the ancient wood object is 11460 years.

Explanation:

We know that half-life for carbon-14 is 5730 years. That means the activity of the sample will decrease to half every 5730 years.

Now we construct the following table:

activity (counts/min) age of the wooden object

40 0

20 5730

10 11460

4 0

3 years ago

gold has a density of 19.32 g/cm3 if you cut a piece of gold in half how much will the density of each piece be?

Rashid [163]

Answer:

19.32 g/cm³

Explanation:

The density will remain the same no matter how many times you cut the gold. The density is g/cm³ or g/mL. Density is essentially how many grams 1 mL of a compound weighs.

7 0

3 years ago

Read 2 more answers

A) The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sampl

diamong [38]

Answer:

For A: The average molecular speed of Ne gas is 553 m/s at the same temperature.

For B: The rate of effusion of $SO_2$ gas is $1.006\times 10^{-3}mol/hr$

Explanation:

For A:

The average molecular speed of the gas is calculated by using the formula:

$V_{gas}=\sqrt{\frac{8RT}{\pi M}}$

$V_{gas}\propto \sqrt{\frac{1}{M}}$

where, M is the molar mass of gas

Forming an equation for the two gases:

$\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}$ .....(1)

Given values:

$V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol$

Plugging values in equation 1:

$\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s$

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

For B:

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

$Rate\propto \frac{1}{\sqrt{M}}$

Where, M is the molar mass of the gas

Forming an equation for the two gases:

$\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}$ .....(2)

Given values:

$Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol$

Plugging values in equation 2:

$\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr$

Hence, the rate of effusion of $SO_2$ gas is $1.006\times 10^{-3}mol/hr$

8 0

3 years ago

a student used 10 mL water instead of 30 mL for extraction of salt from mixture. How may this change the percentage of NaCl extr

Anit [1.1K]

It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:

360 g : 1 L = x g : 30 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = x g : 30 mL

Now, crossing the products:
x · 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000
x = 10.8 g

So, from 30 mL mixture, 10.8 g of NaCl could be extracted.

Let's calculate the same for 10 mL water instead of 30 mL.

360 g : 1 L = x g : 10 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = x g : 10 mL

Now, crossing the products:
x · 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000
x = 3.6 g

So, from 10 mL mixture, 3.6 g of NaCl could be extracted.

Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
3.6/10.8 = 1/3

Therefore, it will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

5 0

3 years ago