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2 years ago

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In a post office, a 3-m long ramp is used to move carts onto a dock that is higher than 1 m. Which describes how the IMA of this

ramp compares with the IMA of the ramp that has a height of 1 m? It is greater. It depends on friction. It is the same. It is less.

1 answer:

pishuonlain [190]2 years ago

6 0

Answer:

well I think the answer is it depends on the friction

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the melting point of scandium fluoride is 1552°c, explain why scandium fluoride has a high melting point.

Greeley [361]

Answer:

Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.

hope it will help you......

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1 year ago

Read 2 more answers

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

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

2 years ago

Based on the products obtained, rank the functional groups (acetamido, amino, and methoxy) in order of increasing ability to act

I am Lyosha [343]

Answer:

Amino >Methoxy > Acetamido

Explanation:

Bromination is of aromatic ring is an electrophilic substitution reaction. The attached functional group to the benzene ring activates or deactivate the aromatic ring towards electrophilic substitution reaction.

The functional group which donates electron to the benzene ring through inductive effect or resonance effect activates the ring towards electrophilic substitution reaction.

The functional group which withdraws electron to the benzene ring through inductive effect or resonance effect deactivates the ring towards electrophilic substitution reaction.

Among given, methoxy and amino are electron donating group. Amino group are stronger electron donating group than methoxy group. Acetamido group because of presence of carbonyl group becomes electron withdrawing group.

Therefore, decreasing order will be as follows:

Amino >Methoxy > Acetamido

7 0

2 years ago

b) If 35 grams of copper (II) chloride react with 20 grams of sodium nitrate, how much sodium chloride can be formed? what is th

charle [14.2K]

Answer:

CuCl₂ is the excessive reactant and NaNO₃ the limiting reactant, 13.7g of NaCl can be formed.

Explanation:

Based on the reaction:

CuCl₂ + 2NaNO₃ → 2NaCl + Cu(NO₃)₂

To solve the problem we must convert the mass of each reactant in order to find limiting reactant as follows:

<em>Moles CuCl₂ -Molar mass: 134.45g/mol-:</em>

35g * (1mol / 134.45g) = 0.26 moles

<em>Moles NaNO₃ -Molar mass: 84.99g/mol-:</em>

20g * (1mol / 84.99g) = 0.235 moles

For a complete reaction of 0.235 moles of NaNO₃ there are required:

0.235 moles NaNO₃ * (1 mol CuCl₂ / 2 mol NaNO₃) = 0.118 moles CuCl₂

As there are 0.26 moles CuCl₂,

<h3>CuCl₂ is the excessive reactant and NaNO₃ the limiting reactant</h3><h3 />

As 2 moles of NaNO₃ produce 2 moles of NaCl, he moles of NaCl produced are: 0.235 moles NaCl. The mass is:

<em>Mass NaCl -Molar mass: 58.44g/mol-:</em>

0.235 moles NaCl * (58.44g / mol) =

<h3>13.7g of NaCl can be formed</h3>

7 0

2 years ago