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torisob [31]
3 years ago
8

I really need the help, come on please, I got a Walmart gift CARD for anyone who can fill this out correctly

Chemistry
1 answer:
Taya2010 [7]3 years ago
6 0

Answer:

1) 1 Na3PO4 + 3 KOH  3 NaOH + 1 K3PO4

2) 1 MgF2 + 1 Li2CO3  1 MgCO3 + 2 LiF

3) 1 P4 + 3 O2  2 P2O3

4) 2 RbNO3 + 1 BeF2  1 Be(NO3)2 + 2 RbF

5) 2 AgNO3 + 1 Cu  1 Cu(NO3)2 + 2 Ag

6) 1 CF4 + 2 Br2  1 CBr4 + 2 F2

7) 2 HCN + 1 CuSO4  1 H2SO4 + 1 Cu(CN)2

Explanation:

8) 1 GaF3 + 3 Cs  3 CsF + 1 Ga

9) 1 BaS + 1 PtF2  1 BaF2 + 1 PtS

10) 1 N2 + 3 H2  2 NH3

11) 2 NaF + 1 Br2  2 NaBr + 1 F2

12) 1 Pb(OH)2 + 2 HCl  2 H2O + 1 PbCl2

13) 2 AlBr3 + 3 K2SO4  6 KBr + 1 Al2(SO4)3

14) 1 CH4 + 2 O2  1 CO2 + 2 H2O

15) 2 Na3PO4 + 3 CaCl2  6 NaCl + 1 Ca3(PO4)2

16) 2 K + 1 Cl2  2 KCl

17) 2 Al + 6 HCl  3 H2 + 2 AlCl3

18) 1 N2 + 3 F2  2 NF3

19) 1 SO2 + 2 Li2Se  1 SSe2 + 2 Li2O

20) 2 NH3 + 1 H2SO4  1 (NH4)2SO4

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They are similarly charged, which is why they repel each other.

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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).

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3 years ago
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zubka84 [21]

Answer: Block # 1 , Block # 4 and

Block # 5 will sink.

Block # 2 and 3 will float.

Explanation: The density of water is equal to 1 g/mL. Any density that is less than the density of water will float. Objects with higher density compared to water will eventually sink.

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What is 108 g +10+5+3
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ANSWER : 108 + 10 = 118

118 + ( 5 + 3 )

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<u>=</u><u> </u><u>1</u><u>2</u><u>6</u><u> </u><u>g</u>

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