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Which of the following is one of the energy conversions taking place in a campfire? Heat to light Heat to kinetic Chemical to li

Blizzard [7]

Answer:Heat to Light

Explanation:

8 0

3 years ago

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PLEASE HELP ME!!!! ILL FIND A WAY TO GET YOU 100 POINTS!!!

Umnica [9.8K]

It’s c had this problem last week

8 0

3 years ago

A 5.00 gram sample of magnesium sulfate hydrate (epsom salt) is heated in the lab to form anhydrous magnesium sulfate. After hea

Ostrovityanka [42]

Answer:

The answer to your question is MgSO₄ 5H₂O

Explanation:

Data

mass of MgSO₄ = 2.86 g

mass of H₂O = 2.14 g (5 - 2.86)

Process

1.- Calculate the molecular mass of the compounds

MgSO₄ = 24 + 32 + (16 x 4) = 120

H₂O = 16 + 2 = 18

2.- Convert the grams obtain to moles

120 g of MgSO₄ --------------- 1 mol

2.8 g ---------------- x

x = (2.8 x 1)/120

x = 0.024 moles

18 g of H₂O --------------------- 1 mol

2.14 g -------------------- x

x = (2.14 x 1)/18

x = 0.119

3.- Divide by the lowest number of moles

MgSO₄ = 0.024/0.024 = 1

H₂O = 0.119/ 0.024 = 5

4.- Write the molecular formula

MgSO₄5H₂O

5 0

3 years ago

How Could Spectroscopy Be Used to Distinguish Between the Following

skelet666 [1.2K]

Spectroscopy be used to distinguish between the following is the compound B has a peak at 3200 – 3500 cm⁻¹ in its IR spectrum.

<h3>What is spectroscopy?</h3>

Spectroscopy is the study of emission or absorption of light. It is used to study the structure of atoms and molecules.

The three types of spectroscopy are:

atomic absorption spectroscopy (AAS)
atomic emission spectroscopy (AES)
atomic fluorescence spectroscopy (AFS)

Thus, the correct option is B, the compound B has a peak at 3200 – 3500 cm⁻¹ in its IR spectrum.

Learn more about spectroscopy

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5 0

2 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

3 years ago