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gregori [183]
2 years ago
7

Task B: Use the following maps to complete the questions:

Chemistry
1 answer:
Oksana_A [137]2 years ago
4 0

Answer:

<h3>1. B</h3><h3>2. A</h3><h3>3. B</h3><h3>4. B</h3><h3>5. C</h3><h3>I HOPE IT HELPS :) 100% sureness</h3>
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An atom has 11 protons, 10 electrons and 13 neutrons. What is the<br> charge for this atom?
Aliun [14]

Answer:

<h3>the charge is +1 </h3>

Explanation:

<h3>as we know nutral atom have equal number of protons and electrons</h3><h3>from the give this element have 11 protons so if it is nutral it must have 11 electrons,but in the question this atom is charged this means it gains or losts certain amount of electrons , this atom has 11 proton and 10 electron from this we can understand this atom dicreases by 1 from its proton, this means it losts one electron .</h3><h3>when an atom lostes electron it's charge become positive with the number of electrons it lostes .</h3><h3>this atom lost 1 electron there fore it have +1 charge and become ion called cathion</h3>

3 0
2 years ago
Which kind of force is a dipole-dipole force?
dexar [7]

Answer:

Explanation:

Types of Attractive Intermolecular Forces. Dipole-dipole forces: electrostatic interactions of permanent dipoles in molecules; includes hydrogen bonding.

This is all I can think of, I hope this has helped you.

-QueenBeauty666-

7 0
3 years ago
Read 2 more answers
The following data were collected for the rate of disappearance of NO in the reaction 2NO(g)+O2(g)→2NO2(g)::
Anit [1.1K]

Answer:

a) The rate law is: v = k[NO]² [O₂]

b) The units are: M⁻² s⁻¹

c) The average value of the constant is: 7.11 x 10³ M⁻² s⁻¹

d) The rate of disappearance of NO is 0.8 M/s

e) The rate of disappearance of O₂ is 0.4 M/s

Explanation:

The experimental rates obtained can be expressed as follows:

v1 = k ([NO]₁)ᵃ ([O₂]₁)ᵇ = 1.41 x 10⁻² M/s

v2 = k ([NO]₂)ᵃ ([O₂]₂)ᵇ = 5.64 x 10⁻² M/s

v3 = k ([NO]₃)ᵃ ([O₂]₃)ᵇ = 1.13 x 10⁻¹ M/s

where:

k = rate constant

[NO]₁ = concentration of NO in experiment 1

[NO]₂ = concentration of NO in experiment 2

[NO]₃ = concentration of NO in experiment 3

[O₂]₁ = concentration of O₂ in experiment 1

[O₂]₂ = concentration of O₂ in experiment 2

[O₂]₃ = concentration of O₂ in experiment 3

a and b = order of the reaction for each reactive respectively.

We can see these equivalences:

[NO]₂ = 2[NO]₁

[O₂]₂ = [O₂]₁

[NO]₃ = [NO]₂

[O₂]₃ = 2[O₂]₂

So, v2 can be written in terms of the concentrations used in experiment 1 replacing [NO]₂ for 2[NO]₁ and [O₂]₂ by [O₂]₁ :

v2 = k (2 [NO]₁)ᵃ ([O₂]₁)ᵇ

If we rationalize v2/v1, we will have:

v2/v1 = k *2ᵃ * ([NO]₁)ᵃ * ([O₂]₁)ᵇ / k * ([NO]₁)ᵃ * ([O₂]₁)ᵇ (the exponent "a" has been distributed)

v2/v1 = 2ᵃ

ln(v2/v1) = a ln2

ln(v2/v1) / ln 2 = a

a = 2

(Please review the logarithmic properties if neccesary)

In the same way, we can find b using the data from experiment 2 and 3 and writting v3 in terms of the concentrations used in experiment 2:

v3/v2 = k ([NO]₂)² * 2ᵇ * ([O₂]₁)ᵇ / k * ([NO]₂)² * ([O₂]₂)ᵇ

v3/v2 = 2ᵇ

ln(v3/v2) = b ln 2

ln(v3/v2) / ln 2 = b

b = 1

Then, the rate law for the reaction is:

<u>v = k[NO]² [O₂]</u>

Since the unit of v is M/s and the product of the concentrations will give a unit of M³, the units of k are:

M/s = k * M³

M/s * M⁻³ = k

<u>M⁻² s⁻¹ = k </u>

To obtain the value of k, we can solve this equation for every experiment:

k = v / [NO]² [O₂]

for experiment 1:

k = 1.41 x 10⁻² M/s / (0.0126 M)² * 0.0125 M = 7.11 x 10³ M⁻² s⁻¹

for experiment 2:

k = 7.11 x 10³ M⁻² s⁻¹

for experiment 3:

k = 7.12 x 10³ M⁻² s⁻¹

The average value of k is then:

(7.11 + 7.11 + 7.12) x 10³ M⁻² s⁻¹ / 3 = <u>7.11 x 10³ M⁻² s⁻¹ </u>

The rate of the reaction when [NO] = 0.0750 M and [O2] =0.0100 M is:

v = k [NO]² [O₂]

The rate of the reaction in terms of the disappearance of NO can be written this way:

v = 1/2(Δ [NO] / Δt) (it is divided by 2 because of the stoichiometric coefficient of NO)

where (Δ [NO] / Δt) is the rate of disappearance of NO.

Then, calculating v with the data provided by the problem:

v = 7.11 x 10³ M⁻² s⁻¹ * (0.0750M)² * 0.0100M = 0.4 M/s

Then, the rate of disappearance of NO will be:

2v = Δ [NO] / Δt = <u>0.8 M/s</u>

The rate of disappearance of O₂ has to be half the rate of disappearance of NO because two moles of NO react with one of O₂. Then Δ [O₂] / Δt = <u>0.4 M/s</u>

With calculations:

v = Δ [O₂] / Δt = 0.4 M/s (since the stoichiometric coefficient is 1, the rate of disappearance of O₂ equals the rate of the reaction).

3 0
3 years ago
at standard pressure, what is the temperature at which a saturated solution of NH4Cl has a concentration of 60g NH4CL/100 g H2O
Nataliya [291]

Answer: Temperature = T, unknown

Saturated Solution, NH4Cl concentration = 60g/100g H2O = 0.6g NH4Cl/g H2O

Assume density of H2O = 1 g/ml

m = 0.6g NH4Cl/g H2O / 1 g/ml

m = 0.6g NH4Cl/ml

See the table of saturated solutions and identify the temperature at which the concentration of NH4Cl is 60g/100g H2O.

Explanation:  The line on the graph on reference table G indicates a saturated solution of NH4CL as a concentration of 60. g NH4 Cl/100. g H2O

5 0
2 years ago
One mole of water weights?
Gemiola [76]

Answer:

c 18.0ml

Explanation:

The average mass of one H2O molecule is 18.02 amu. The number of atoms is an exact number, the number of mole is an exact number; they do not affect the number of significant figures. The average mass of one mole of H2O is 18.02 grams. This is stated: the molar mass of water is 18.02 g/mol.

8 0
3 years ago
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