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mariarad [96]
2 years ago
12

A 1.00-degree increase on the Celsius scale is equivalent to a 1.80-degree increase on the Fahrenheit scale. The temperature of

a fluid increases by 48.0°C. What is this increase in degrees Fahrenheit? 49.8°F 48.0°F 26.7°F 86.4°F
Chemistry
2 answers:
muminat2 years ago
8 0

Answer: last optiion, 86.4°F

Explanation:

1) Use the symbol Δ for increase

Given: Δ1.00 °C = Δ1.80°F, and Δ48.0°C, find Δ°F = x.

2) Proportion:

Δ1.00 °C / Δ1.80°F = Δ48.0°C / x.

Solve for x: x = Δ48.0°C × Δ 1.8 °F / Δ1°C = 86.4°F

Conclusion: the increase equals 86.4°F


oksian1 [2.3K]2 years ago
6 0
1.00 degree increase<span> on the </span>Celsius scale<span> is </span>equivalent<span> to a </span>1.80 degree increase<span> on the</span>Fahrenheit scale<span>. If a </span>temperature increases<span> by </span>48.0°C<span>, </span>
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Net Primary Productivity ... the amount of biomass present in an ecosystem at a particular time .... Explain why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.
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A student adds solid KCl to water in a flask. The flask is sealed with a stopper and thoroughly shaken until no more solid KCl d
AlladinOne [14]

Answer:

Option (A) saturated and is at equilibrium with the solid KCl

Explanation:

A saturated solution is a solution which can not dissolve more solute in the solution.

From the question given above, we can see that the solution is saturated as it can not further dissolve any more KCl as some KCl is still visible in the flask.

Equilibrium is attained in a chemical reaction when there is no observable change in the reaction system with time. Now, observing the question given we can see that there is no change in flask as some KCl is still visible even after thorough shaking. This simply implies that the solution is in equilibrium with the KCl solid as no further dissolution occurs.

4 0
2 years ago
Paige heated 3.00 g mercury (II) oxide (HgO, 216.59 g/mol) to form mercury (Hg, 200.59 g/mol) and oxygen (O2, 32.00 g/mol). She
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the balanced chemical equation for decomposition of HgO is as follows

2HgO --> 2Hg + O₂

stoichiometry of HgO to O₂ is 2:1

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according to stoichiometry of reaction -

number of O₂ moles formed = 0.0139 mol/ 2 = 0.00695 mol

mass of O₂ to be formed - 0.00695 mol x 32.00 g/mol = 0.2224 g

but the actual yield = 0.195 g

percent yield = actual yield / theoretical yield x 100 %

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8 1
2 years ago
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Which contain covalent bonds?
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Determine net ionic equations, if any, occuring when aqueous solutions of the following reactants are mixed. Select "True" or "F
nirvana33 [79]

<u>Answer:</u>

<u>For 1:</u> The correct answer is False.

<u>For 2:</u> The correct answer is True.

<u>For 3:</u> The correct answer is True.

<u>For 4:</u> The correct answer is False.

<u>For 5:</u> The correct answer is True.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.  If no net ionic equation is formed, it is said that no reaction has occurred.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. Solids, liquids and gases do not exist as ions.

  • <u>For 1:</u> Lead(II) nitrate and sodium chloride

The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:

Pb(NO_3)_2(aq.)+2NaCl(aq.)\rightarrow PbCl_2(s)+2NaNO_3(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+2Na^+(aq.)+2NO_3^-(aq.)

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is False.

  • <u>For 2:</u> Sodium bromide and hydrochloric acid

The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:

NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)

Ionic form of the above equation follows:

Na^{+}(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)

There are no spectator ions in the equation. So, the above reaction is the net ionic equation.

Hence, the correct answer is True.

  • <u>For 3:</u> Nickel (II) chloride and lead(II) nitrate

The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:

Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+Ni^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+Ni^{2+}(aq.)+2NO_3^-(aq.)

As, nickel and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is True.

  • <u>For 4:</u> Magnesium chloride and sodium hydroxide

The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:

MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)

Ionic form of the above equation follows:

Mg^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Mg^{2+}(aq.)+OH^-(aq.)\rightarrow Mg(OH)_2(s)

Hence, the correct answer is False.

  • <u>For 5:</u> Ammonium sulfate and barium nitrate

The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:

(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)

Ionic form of the above equation follows:

2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ba^{2+}(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)

Hence, the correct answer is True.

3 0
2 years ago
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