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2 years ago

A substance with a mass of 50.0 g has a temperature change of 40.0

Chemistry

2 answers:

grigory [225]2 years ago

7 0

Answer:

Mass=50.0g

H=670J

change in temperature=40

using. c=h÷m×change in temperature

c=670÷50×40

C=670÷2000

C=0.335jkg-1k-1

Alex2 years ago

6 0

Answer: The specific heat of a substance with a mass of 50. 0g and a temperature change of 40. 0 degrees whne it absorbs 670 J of heat is 337 J/kg. K

Explanation:

Specific heat of a substance is said to be the energy required for an increase in temperature by 1°C in the unit of J/Kg. K

Formula of Specific heat:

Q = mcΔθ

Q represents the quantity of heat

m represents the mass of the substance in Kilograms = 50. 0g÷ 1000 = 0.05kg

c represents the specific heat capacity of the substance = ?

θ represents the temperature change = 40° C

Specific heat, c = Q ÷ mΔθ

= 670 ÷ 0.05 × 40

= 670 ÷ 2

= 335 J/ Kg. K

Therefore, the specific heat of the substance is 335 J/kg. K

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Explanation:

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2 years ago

In chemistry what is Faraday's law

Eduardwww [97]

Answer:

Faraday’s – First Law of Electrolysis

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Faraday’s – Second Law of Electrolysis

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From these laws of electrolysis, we can deduce that the amount of electricity needed for oxidation-reduction depends on the stoichiometry of the electrode reaction.

The product of an electrolytic reaction depends on the nature of the material being electrolysed and the type of electrodes used. In the case of an inert electrode such as platinum or gold, the electrode does not participate in the chemical reaction and acts only as a source or sink for electrons. While, in the case of a reactive electrode, the electrode participates in the reaction.

Hence, different products are obtained for electrolysis in the case of reactive and inert electrodes. Oxidizing and reducing species present in the electrolytic cell and their standard electrode potential too, affect the products of electrolysis.

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2 years ago

Read 2 more answers

Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int

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Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.

Million Gallons per day $1 MGD = 3785411.8 litre/day = 3785411.8 L/d$

$F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d$

We have one flow of wastewater released into a stream.

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:

$C_f = \frac{F1*C1 +F2*C2}{F1 +F2}$

Replacing every value in L/d and mg/L

$C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L$

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).

We have the total flow F3 = F1 + F2 and the final concentration $C_f$ . It is required to calculate per day, let's take a time of t = 1 day.

$F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg$

After that, mg are converted to pounds.

$M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb$

b) A total of 137.6 pounds pass a given spot downstream per day.

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3 years ago

Extreme temperature changes or increased moisture speed up the weathering rate.

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3 years ago

How many atoms are there in 7.2 mol neon?

rosijanka [135]

Answer:

A) 4.3 × 10²⁴ atoms

Explanation:

Step 1: Given data

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Step 2: Calculate the number of atoms present in 7.2 moles of neon

In order to convert moles to toms, we need a conversion factor. In this case, we will use Avogadro's number: there are 6.02 × 10²³ neon atoms in 1 mole of neon atoms.

7.2 mol × 6.02 × 10²³ atoms/mol = 4.3 × 10²⁴ atoms

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2 years ago