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1 year ago

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Chemistry

1 answer:

Diano4ka-milaya [45]1 year ago

3 0

Answer:

The advantage of net ionic equations is that they show only those species that are directly involved in the reaction

Explanation:

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A beaker contains a 25 ml solution of an unknown monoprotic acid that reacts in a 1:1 stoichiometric ratio with naoh. titrate th

yuradex [85]

Given:

Volume of the unknown monoprotic acid (HA) = 25 ml

To determine:

The concentration of the acid HA

Explanation:

The titration reaction can be represented as-

HA + NaOH → Na⁺A⁻ + H₂O

As per stoichiometry: 1 mole of HA reacts with 1 mole of NaOH

At equivalence point-

moles of HA = moles of NaOH

For a known concentration and volume of added NaOH we have:

moles of NaOH = M(NaOH) * V(NaOH)

Thus, the concentration of the unknown 25 ml (0.025 L) of HA would be-

Molarity of HA = moles of HA/Vol of HA

Molarity of HA = M(NaOH)*V(NaOH)/0.025 L

5 0

1 year ago

When magnesium ribbon is placed in sulfuric acid, a chemical reaction takes place and heat energy is given off. Liang and Naraba

kolezko [41]

Answer:

this is the answer

I hope it helps you although

6 0

1 year ago

A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.

Sonja [21]

Answer:

Equilibrium constant for $PCl_5$ is 0.5

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

Explanation:

$PCl_5$ dissociates as follows:

$PCl_5 \rightleftharpoons PCl_3+Cl_2$

initial 0.72 mol 0 0

at eq. 0.72 - 0.40 0.40 0.40

Expression for the equilibrium constant is as follows:

$k=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Substitute the values in the above formula to calculate equilibrium constant as follows:

$k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5$

Therefore, equilibrium constant for $PCl_5$ is 0.5

Now calculate the equilibrium constant for decomposition of $NO_2$

It is given that $3.3 \times 10^{-3} \%$ is decomposed.

$NO_2$ decomposes as follows:

$2NO_2 \rightleftharpoons 2NO + O_2$

initial 1.0 M 0 0

at eq. concentration of $NO_2$ is:

$[NO_2]_{eq}=1-(0.000066) = 0.999934\ M$

$[NO]_{eq}=6.6 \times 10^{-5}\ M$

$[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M$

Expression for equilibrium constant is as follows:

$K=\frac{[NO]^2[O_2]}{[NO_2]^2}$

Substitute the values in the above expression

$K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}$

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

8 0

1 year ago

The three‑dimensional structure of a generic molecule is given. Identify the axial and equatorial atoms in the three‑dimensional

Dima020 [189]

Answer:

Explanation:

CHECK THE ATTACHMENT FOR THE COMPLETE QUESTION AND THE DETAILED EXPLANATION

NOTE:

Equatorial atoms are referred to atoms that are attached to carbons in the cyclohexane ring which is found at the equator of the ring.

Axial atoms are atoms that exist in a bond which is parallel to the axis of the ring in cyclohexane

4 0

1 year ago

What volume of a 0.716 m kbr solution is needed to provide 30.5 g of kbr?

Jlenok [28]

Answer is: volume of KBr is 357 mL.
c(KBr) = 0,716 M = 0,716 mol/L.
m(KBr) = 30,5 g.
n(KBr) = m(KBr) ÷ M(KBr).
n(KBr) = 30,5 g ÷ 119 g/mol.
n(KBr) = 0,256 mol.
V(KBr) = n(KBr) ÷ c(KBr).
V(KBr) = 0,256 mol ÷ 0,716 mol/L.
V(KBr) = 0,357 L · 1000 mL/L = 357 mL.

7 0

1 year ago

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