Over time coral buried by sediments can turn into

Create a list of 5 potential jobs that students of chemistry can obtain. Which job appeals to you the most?

Korvikt [17]

Answer:

cleaning

someone can make sure everyone is on task

the data taker

the person who measures everything

the person who makes sure yall have all the material for the experiment

Explanation:

hope this was what u were looking for!

3 0

3 years ago

Read 2 more answers

QUESTION 1

Andrews [41]

Answer:

It appears all of them are correct
Smaller drops will scatter around while maintaining their shape
Larger drop will spread stick to the surface.
and drops close together will join to form on gigantic sphere(but not perfect sphere)
and if they are in the sun, they'll eventually evaporate

Explanation:

7 0

2 years ago

C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate constant, ????

Yanka [14]

Answer:

Rate constant at 725 K is $5.2\times 10^{-4}s^{-1}$

Explanation:

According to Arrhenius equation for a reaction-

$ln(\frac{k_{2}}{k_{1}})=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

where $k_{2}$ and $k_{1}$ are rate constants of reaction at $T_{2}$ and $T_{1}$ temperatures (in kelvin) respectively.

$E_{a}$ is activation energy of reaction.

Here $T_{1}$ = 600 K , $k_{1}$ = $6.1\times 10^{-8}s^{-1}$

$T_{2}$ = 725 K, $E_{a}$ = 262 kJ/mol and R = 8.314 J/(mol.K)

So plugin all the values in the above equation-

$ln(\frac{k_{2}}{6.1\times 10^{-8}s^{-1}})=\frac{262\times 10^{3}J/mol}{8.314J/(mol.K)}\times (\frac{1}{600K}-\frac{1}{725K})$

So, $k_{2}$ = $5.2\times 10^{-4}s^{-1}$

Hence rate constant at 725 K is $5.2\times 10^{-4}s^{-1}$

7 0

4 years ago

Sodium hydroxide reacts with carbondioxide as follows: 2 naoh(s) + co2 (g) → na2co3 (s) + h2o(l) which reagent is the limiting r

tensa zangetsu [6.8K]

1) Balanced chemical equation:

2 NaOH(s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

2) Mole ratio

2 mole NaOH : 1 mol CO₂ : 1 mol Na₂CO₃ : 1 mol H₂O

3) Limiting reactant

1.85 mol NaOH / 1.00 mol CO₂ < 2 mole NaOH / 1 mol CO₂ ⇒ NaOH is not enough to react with 1.00 mole of CO₂ (more NaOH is needed), so NaOH is the limiting reactant.

Answer: the limiting reactant is NaOH

4) How many moles of sodium carbonate can be produced?

Set the proportion relation using the limiting reactant (the one that reacts completely) and the theoretical mole ratio.

2 mole NaOH / 1mol Na₂CO₃ = 1.85 mol NaOH / x

x = 0.925 mol Na₂CO₃

Answer: 0.925 moles Na₂CO₃

5) How many moles of the excess reactant remain after the completion of the reaction?

The excess reactant is CO₂.

The amount of CO₂ consumed is calculated with a proportion:

2 mol NaOH / 1 mol CO₂ = 1.85 mol NaOH / x

⇒ x = 1.85 / 2 = 0.925mol CO₂

The amount remaining is the original amount less the amount that reacted = 1 mol - 0.925 mol = 0.075 mol.

Answer: 0.075 mol CO₂

5 0

3 years ago

Calculate the mass (in grams) of methylene bluecrystals that you must weigh in order to make 100.0mL of 1.25 × 10-5mol/L methyle

mario62 [17]

<u>Answer:</u> The mass of methylene blue that must be weighed is $3.99\times 10^{-4}g$

<u>Explanation:</u>

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = $1.25\times 10^{-5}M$

Molar mass of methylene blue = 319.85 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$1.25\times 10^{-5}M=\frac{\text{Mass of methylene blue}\times 1000}{319.85\times 100.0}\\\\\text{Mass of methylene blue}=\frac{1.25\times 10^{-5}\times 319.85\times 100.0}{1000}=3.99\times 10^{-4}g$

Hence, the mass of methylene blue that must be weighed is $3.99\times 10^{-4}g$

4 0

3 years ago