Water covers over 70% of planet Earth and most of the water is salt water. Which

How could you use a model to show the cause-and-effect relationship between Earth's rotation and the apparent motion of the star

Masteriza [31]

Question:

How could you use a model to show the cause-and-effect relationship between Earth's rotation and the apparent motion of the stars across the night sky?

Answer:

Gravity? or density because of the pull from the sun.

7 0

3 years ago

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What is the ratio of effusion rates of krypton and neon at the same temperature and pressure?

egoroff_w [7]

The formula we use would be the graham's law. We do as follows:

E_Kr / E_Ne = sqrt ( M_Ne / M_Kr)

= sqrt ( 20.1797 g/mol / 83.798 g/mol ) 
= sqrt (0.24081) 
= 0.4907

Hope this answers the question. Have a nice day.

7 0

3 years ago

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A 0.100 M solution of bromoacetic acid (BrCH2COOH) is 13.2% ionized. Calculate [H+]? Calculate BrCH2COO^ - ? Calculate BrCH2COOH

Natali [406]

Answer: The concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M

Explanation:

We are given:

Molarity of bromoacetic acid = 0.100 M

Percent of ionization = 13.2 %

The chemical equation for the ionization of bromoacetic acid follows:

$BrCH_2COOH\rightarrow BrCH_2COO^-+H^+$

1 mole of bromoacetic acid produces 1 mole of bromoacetate ion and 1 mole of hydrogen ion

Molarity of hydrogen ion = 13.2 % of 0.100 = $\frac{13.2}{100}\times 0.100=0.0132M$

Molarity of bromoacetate ion = molarity of hydrogen ion = 0.0132 M

Molarity of bromoacetic acid = Molarity of solution - Molarity of ionized substance

Molarity of bromoacetic acid = 0.100 - 0.0132 = 0.0868 M

Hence, the concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M

8 0

3 years ago

An alkyne with the molecular formula C5H8 was reduced with H2 and Lindlar's catalyst. Hydroboration-oxidation of the resulting a

irinina [24]

Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.

The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.

When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.

Learn more: brainly.com/question/2510654

6 0

3 years ago

4. Magnesium and oxygen undergo a chemical reaction to

erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
$\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}$

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

Convert grams of MgO to moles of MgO.
Moles of MgO to moles of O₂
And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

$\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

$\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

3 0

2 years ago