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3 years ago

What is the correct order of processes in the accretion of planetesimals?

Chemistry

1 answer:

icang [17]3 years ago

4 0

Answer:

A). Collapse, condensation, accretion, clearing.

Explanation:

Planetesimals are described as the small solid astronomical objects which are the building blocks for the formation of solid planets. Accretion is described as the process in which the solid astronomical objects cluster to form larger objects that eventually leads to the formation of planets.

As per the question, the correct order of processes that are followed in this accretion process would be a collapse, condensation, accretion, clearing. First occurs the collapse of a nebula followed by condensation of cosmic dust grains. After this, the accession or accumulation of solid parts takes place to form larger objects and this is immediately followed by the clearing of debris and dust to allow planets to form. Thus, option A is the correct answer.

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Which of the following is an adaptation that helps protect the animal?

saveliy_v [14]

C ( Porcupines use there’s sharp quills to defend themselves from larger predators!)

6 0

2 years ago

The following reaction: HF(aq) <--> H+(aq) + F-(aq) has an equilbirum constant (K) value of 7.2 x 10-4. This means that th

andriy [413]

Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.

<h3>What is the equilibrium constant?</h3>

In a reaction, we can judge using the value of the equilibrium constant weather or not the reaction moves on to completion. If the reaction moves up to completion, it the follows that the value of the equilibrium constant ought to be large.

On the other hand, when we have a case that the equilibrium constant is small and is not so large, then the reaction does not proceed essentially to completion.

Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.

Learn more about equilibrium constant:brainly.com/question/10038290

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6 0

1 year ago

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

2 years ago

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago

What is the molarity of 20.2 g of potassium nitrate, KNO3, in enough water to make 250.0 mL of solution

KATRIN_1 [288]

Answer:

0.8M

Explanation:

CM=n/V

8 0

2 years ago