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2 years ago

Plz help... answer this question!

Chemistry

1 answer:

Mkey [24]2 years ago

5 0

Answer:

Explanation:

Because element is made up of only one type of atoms and figure b show only one type of atoms.

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What is the lowest point of a wave called?<br><br> Amplitude<br> Crest<br> Frequency<br> Trough

zavuch27 [327]

<h3>Answer:</h3>

Trough

<h3>Explanation:</h3>

We need to answer the following questions

What is a wave?

A wave refers to a transmission of a disturbance from one point to another. It involves transfer of energy from the source to other points.

What are the major types of wave?

Waves may be classified as longitudinal waves or transverse wave based on the vibration of particles relative to the direction of transmission.

What is a transverse wave?

A transverse wave is a wave in which the vibration of particles is perpendicular to the direction of transmission of the wave.
This type of wave contains highest points known as crests and lowest points known as troughs.

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3 years ago

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musickatia [10]

Answer:

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3 years ago

Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO

sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

CaO : CaCO₃

1 : 1

0.26 : 0.26

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ = 26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

7 0

3 years ago

For the Earth's atmosphere, section number two in the pie graph BEST represents the percentage of A) argon B) oxygen C) nitrogen

larisa86 [58]

-<u><em>Oxygen</em></u>

According to Google these are the percentages of the <em>Earths Atmosphere</em>

<em>1</em> 78% - Nitrogen

<u>2</u> 21% - Oxygen

<em>3</em> 0.9% - Argon

<em>4 </em>0.3 - Carbon Dioxide with very small percentage of other elements.

3 0

3 years ago

A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

3 0

3 years ago