Glucose, C₂H12O6, is a product of photosynthesis.

Why was Niels Bohr’s atomic model superior to all the earlier models?

sergeinik [125]

Answer:

B

Explanation:

The Niels Bohr's atomic model superior to all the earlier models is because it showed how the electron could orbit the nucleus without falling into it

7 0

3 years ago

Witch of these is an example of an environment changing slowly

pshichka [43]

It is important to note that whenever the physical environment is changed animals and plants have to adapt or they will go extinct. Overtime environments change in some instances is due to the slow movement of the continents which can take millions of years, Earth's climate also causes changes in the environment over a long period of time. Additionally, some changes can also be caused by people since people can change forests into farms or intro concrete streets as such.

6 0

3 years ago

What is the density of an object with a mass of 145.8g and an volume of 91.75 mL?

Over [174]

Answer:

<h2>Density = 1.6 g/mL</h2>

Explanation:

Density of a substance can be found by using the formula

$Density = \frac{mass}{volume}$

From the question

mass = 145.8 g

volume = 91.75 mL

Substitute the values into the above formula and solve for the Density

That's

$Density = \frac{145.8}{91.75}$

= 1.5891

We have the final answer as

Density = 1.6 g/mL

Hope this helps you

4 0

3 years ago

For the chemical reaction:

Julli [10]

Answer:

Volume of ammonia produced = 398.7 dm³

Explanation:

Given data:

Volume of N₂ = 200 dm³

Pressure and temperature = standard

Volume of ammonia produced = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Number of moles of N₂:

PV = nRT

1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K

n = 200 atm.L /22.41 atm.L/mol

n = 8.9 mol

Now we will compare the moles of ammonia and nitrogen.

N₂ : NH₃

1 : 2

8.9 : 2/1×8.9 = 17.8 mol

Volume of ammonia:

1 mole of any gas occupy 22.4 dm³ volume

17.8 mol ×22.4 dm³/1 mol = 398.7 dm³

4 0

3 years ago

"Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

Alex17521 [72]

Answer:

$\boxed{\text{0.50 mol/L}}$

Explanation:

The balanced equation is

2COF₂ ⇌ CO₂+CF₄; Kc = 9.00

1. Set up an ICE table

$\begin{array}{ccccc}\rm 2COF_{2} & \, \rightleftharpoons \, & \rm CO_{2} & +&\rm CF_{4}\\2.00& & 0& & 0\\-x& & +x & & +x\\2.00 - x& & x & &x \\\end{array}$

2. Solve for x

$K_{c} = \dfrac{[\rm CO][ \rm CF_{4}]}{[\rm COF_{2}]^{2}} = 9.00\\\\\begin{array}{rcl}\dfrac{x^{2}}{(2.00 - x)^{2}} & = & 9.00\\\dfrac{x}{2.00 - x} & = & 3.00\\x & = &3.00(2.00 - x)\\x & = & 6.00 - 3.00x\\4.00x & = & 6.00\\x & = & \mathbf{1.50}\\\end{array}$

3. Calculate the equilibrium concentration of COF₂

c = (2.00 - x) mol·L⁻¹ = (2.00 - 1.50) mol·L⁻¹ = 0.50 mol

$\text{The equilibrium concentration of COF$_{2}$ at equilibrium is $\boxed{\textbf{0.50 mol/L}}$}$

Check:

$\begin{array}{rcl}\dfrac{1.50^{2}}{0.50^{2}} & = & 9.00\\\\\dfrac{2.25}{0.25}& = & 9.00\\\\9.00 & = & 9.00\\\end{array}$

OK.

5 0

4 years ago