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2 years ago

Brainlist for best answer!!! Why will the whole ecosystem be affected by just one change?

Chemistry

2 answers:

Naddika [18.5K]2 years ago

7 0

Answer:

Critical coordinate drivers incorporate living space alter, climate alter, obtrusive species, over exploitation, and contamination. Most of the coordinate drivers of corruption in biological systems and biodiversity right now stay steady or are developing in escalated in most biological systems.

Explanation:

juin [17]2 years ago

5 0

Wind, rain, predation and earthquakes are all examples of natural processes which impact an ecosystem. Humans also affect ecosystems by reducing habitat, over-hunting, broadcasting pesticides or fertilizers, and other influences. The line between natural and human caused effects often blurs. Brainiest please!!!

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The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to conve

coldgirl [10]

Answer:

80.27%

Explanation:

Let's consider the following balanced equation.

2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)

First, we have to calculate the moles of Sn²⁺ that react.

$\frac{0.1015molSn^{2+} }{1L} .13.28 \times 10^{-3} L=1.348\times 10^{-3}molSn^{2+}$

We also know the following relations:

According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
The molar mass of Fe is 55.84 g/mol.

Then, for 1.348 × 10⁻3 moles of Sn²⁺:

$1.348\times 10^{-3}molSn^{2+}.\frac{2molFe^{3+} }{1molSn^{2+} } .\frac{1molFe}{1molFe^{3+} } .\frac{55.84gFe}{1molFe} =0.1505gFe$

If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:

$\frac{0.1505g}{0.1875g} \times 100 \% = 80.27\%$

5 0

2 years ago

The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other plati

Alinara [238K]

Here we have to identify the metal powder by the given disposal.

The identification of Zinc can be done by 1 m nitric acid (HNO₃) and Ni(NO₃)₂ which will produce hydrogen gas by reaction and displacement reaction as shown below.

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

The identification of lead can be done by the reaction with 1 m nitric acid (HNO₃) which produces lead nitrate.

The reaction is-

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

The identification of platinum can be done by the reaction with all the given disposal as it will not react with any of the compound.

1. Identification of Zinc (Zn):

(a) Zn metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than zinc.

Zn + NaNO₃ = No reaction

(b) Zn metal will react with 1 m HNO₃ to form hydrogen gas. The reaction is:

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

(c) Zn will react with nickel nitrate [Ni(NO₃)₂] because it may only cause displacement reaction the reduction potential of Zn²⁺/Zn (-0.76) is less than that of Ni²⁺/Ni (-0.23). Thus the reaction will be:

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

2. Identification of lead (Pb):

(a) Pb metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pb.

Pb + NaNO₃ = No reaction

(b) Pb reacts with HNO₃ to form lead nitrate. The reaction is:

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

(c) The standard reduction potential of Pb²⁺/Pb is more than nickel Ni²⁺/Ni thus there will be no reaction between Pb and NI(NO₃)₂.

Pb + Ni(NO₃)₂ = No reaction.

3. Identification of platinum (Pt)

(a) Pt metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pt.

Pt + NaNO₃ = No reaction.

(b) The standard reduction potential of Pt²⁺/Pt is so high (+1.188) thus there will be no reaction with HNO₃.

Pt + HNO₃ = No reaction

(c) The standard reduction potential of Pt²⁺/Pt is more than nickel Ni²⁺/Ni thus there will be no reaction between Pt and Ni(NO₃)₂.

Pt + Ni(NO₃)₂ = No reaction.

8 0

2 years ago

In a study of the following reaction at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0

sp2606 [1]

Answer:

The value of $K_p$ for this reaction at 1200 K is 4.066.

Explanation:

Partial pressure of water vapor at equilibrium = $p^o_{H_2O}=15.0 Torr$

Partial pressure of hydrogen gas at equilibrium = $p^o_{H_2}=?$

Total pressure of the system at equilibrium P = 36.3 Torr

Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:

$P=p^o_{H_2O}+p^o_{H_2}$

$p^o_{H_2}=P-p^o_{H_2O}=36.3 Torr- 15.0 Torr = 21.3 Torr$

$3 Fe(s) 4 H_2O(g)\rightleftharpoons Fe_3O_4(s) 4 H_2(g)$

The expression of $K_p$ is given by:

$K_p=\frac{(p^o_{H_2})^4}{(p^o_{H_2O})^4}$

$K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}=4.066$

The value of $K_p$ for this reaction at 1200 K is 4.066.

6 0

2 years ago

Which of these phrases accurately describe disproportionation? Check all that apply.

Olin [163]

Answer:

A single compound is simultaneously oxidized and reduced.

Explanation:

In chemistry, disproportionation is a simultaneous oxidation and reduction of a single chemical specie.

What this means is that; in a disproportionation reaction, only one compound is both oxidized and reduced. This implies that two products are formed during disproportionation. One is the oxidized product while the other is the reduced product.

Consider the disproportionation of CuCl shown below;

2CuCl -----> CuCl2 + Cu

Here, CuCl2 is the oxidized product while Cu is the reduced product.

7 0

1 year ago

Read 2 more answers

A 0.2688 g sample of a monoprotic acid neutralized 16.4 mL of 0.08133 M KOH solution. Calculate the molar mass of the acid

ivanzaharov [21]

Answer:

202 g/mol

Explanation:

Let's consider the neutralization between a generic monoprotic acid and KOH.

HA + KOH → KA + H₂O

The moles of KOH that reacted are:

0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol

The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.

1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:

0.2688 g/1.33 × 10⁻³ mol = 202 g/mol

3 0

2 years ago