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Ksenya-84 [330]
3 years ago
6

Which formula can be used to calculate the molar mass of hydrogen peroxide (H2O2)? (5 points)

Chemistry
1 answer:
ratelena [41]3 years ago
7 0

<em>answer:</em><em> </em><em>option </em><em>d </em><em>(</em><em>2</em><em>×</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>mass </em><em>of </em><em>H </em><em>+</em><em>2</em><em>×</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>mass </em><em>of </em><em>O</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

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The ground state of an electron is the least stable energy state of an atom
kari74 [83]

Answer:

electron configuration

Explanation:

The arrangement of electrons in the atomic orbitals of an atom is called the electron configuration. Electron configurations can be determined using a periodic table.

6 0
3 years ago
A gas occupies 40.0 mL at 150 K. What volume does it occupy at 300 K,
nlexa [21]

Answer:

.08 L or 80 ml

Explanation:

Use the equation V/t = V/t.

.04L / 150K = V / 300K

.04 / 150 * 300 = V

.08 L or 80 ml

6 0
3 years ago
Can sedimentation and decantation be used for all types of mixtures? Explain
Bumek [7]

Explanation:

1. Sedimentation and decantation cannot be used for all types of mixtures.

Decantation is a separation technique in which is used to separate immiscible liquids or mixtures containing liquid and solids within them.

In decantation, gravity is used to bring the denser materials to settle at the bottom.

For homogenous mixtures, it is not possible to use decantation. A solution of sugar and water will not decant.

2. Yes, mass of an object reduces the settling time of such object in a mixture.

The higher the mass, the faster the rate of settling. Also, as we know, mass is directly proportional to density. A body with a high density will settle faster in solution.

4 0
3 years ago
How much energy does it take to boil 500.g of water at its boiling point?
Tatiana [17]

Answer:

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Explanation:

7 0
3 years ago
The rate constant for this second‑order reaction is 0.610 M − 1 ⋅ s − 1 0.610 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product
Darya [45]

Answer: It takes 3.120 seconds for the concentration of  A to decrease from 0.860 M to 0.260 M.

Explanation:

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

k = rate constant = 0.610M^{-1}s^{-1}

a_0 = initial concentration = 0.860 M

a= concentration left after time t = 0.260 M

\frac{1}{0.260}=0.860\times t+\frac{1}{0.860}

t=3.120s

Thus it takes 3.120 seconds for the concentration of  A to decrease from 0.860 M to 0.260 M.

8 0
3 years ago
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