Will give brainliest

Which two from this group have an identical number of energy levels?

storchak [24]

Answer: C

Explanation: Took the test

5 0

3 years ago

5. How many grams are in 2.30 moles of silicon dioxide?

tino4ka555 [31]

Answer:

138.19388999999998 grams of silicon dioxide

7 0

2 years ago

Consider the aldol condensation between dibenzyl ketone and benzil in strong base. Calculate the molar masses of the reactants a

Sati [7]

Answer:

a) 210.3 g/mol

b) 210.2 g/mol

c) 384.5 g/mol

Explanation:

First step we will calculate the molar masses of ; carbon atom, hydrogen atom and oxygen atom in each .

Molar mass of dibenzyl ketone

Molar mass of dibenzyl ketone = ∑ molar masses of atoms in dibenzyl ketone

= carbon( 15 ) = 15 ( 12.0107 ) + oxygen ( 14 ) = 1 ( 15.999 ) + hydrogen(14) =14(1.00784)

= 210.26926 ≈ 210.3 g/mol

Molar mass of benzil

Molar mass of Benzil = ∑ molar masses of atoms in Benzil

= carbon( 14) = 14(12.0107) + oxygen(2) = 2 ( 15.999) + hydrogen(10) =10(1.00784)

= 210.2262 ≈ 210.2 g/mol

Molar mass of 2,3,4,5-tetraphenylcyclopentadienone

Molar mass = ∑ molar masses of atoms

= carbon ( 29) = 29(12.0107) + oxygen (1) = 1( 15.999 ) + hydrogen(20) = 20(1.00784 )

≈ 384.5 g/mol

6 0

3 years ago

What happens to the pH of a solution is the concentration of hydronium ions decrease

insens350 [35]

If the hydronium concentration decreases, the pH increases, resulting in a solution that is less acidic and more basic

5 0

3 years ago

Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)

allsm [11]

Answer:

$\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}$

Explanation:

1. Density from mass and volume

$\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}$

2. Volume from density and mass

$V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}$

3. Mass from density and volume

$\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}$

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water = 12.8 mL

Volume of object = 11.8 mL

$\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}$

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

6 0

3 years ago