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weqwewe [10]
4 years ago
9

Which scientist's experiments proved the existence of the nucleus? Niels Bohr's J.J. Thomson's Robert Millikan's Ernest Rutherfo

rd's
Chemistry
1 answer:
bezimeni [28]4 years ago
5 0

Answer: Ernest Rutherford.

Explanation:

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You are asked to bring the pH of 0.500L of 0.550 M NH4Cl to 7.00. Which of the following solutions would you use: 12.0 M HCl or
ch4aika [34]
<span>1 drop is approximately 0.05mL. Since 0.500L of 0.550M NH4Cl contains 0.275mol of substance (calculated by using c=n/V formula), equal amount of substance of NH3 is needed to neutralize this solution (since pH of 7 is neutral solution). Thus, we need 0.0275L of NH3, i.e. around 550 drops.</span>
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Burning of wood is physical or chemical change​
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chemical change

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3 years ago
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Lena [83]

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3 years ago
Explain how a motorcar engine produces nitrogen oxides
olya-2409 [2.1K]

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4 0
3 years ago
Read 2 more answers
My swimming pool is rectangular (16 feet by 34 feet) and has a depth of 6 feet. Lets imagine that my pool water is full to the t
Reil [10]

Answer:

Number of moles of photons required = 5.04 × 10⁴ moles

Explanation:

The energy of a photon can be calculated from Planck's equation E = hc/λ

Where h = 6.63 × 10-³⁴ Js and c, the velocity of light = 3.0 × 10⁸ m/s

Energy of one mole of photons = N₀ × hc/λ

wavelength of photon, λ = 520 nm = 5.20 × 10-⁷ m

Energy of one mole of photons = 6.02 × 10²³ × 6.63 × 10−³⁴ × 3 × 10⁸/5.20 × 10-⁷

Energy of one mole of photons = 2.30 × 10⁵ J/mol

Energy required to raise the temperature of a given mass of a substance, E = mcΔT

Where m is mass of substance,  c is specific heat capacity,  ΔT is temperature difference

Mass ofnwternin the pool = volume × density

Volume of water = Volume of swimming pool

Volume of water = 16 × 34 × 6 ft³ = 3264 ft³

1 ft³ = 28316.8 cm³; 3264 ft³ = 28316.8 × 3264 = 92426035.2 cm³

Density of water = 1 g/cm³

Mass of water = 92426035.2 cm³ × 1 g/cm³ = 92426035.2g

ΔT = 80°C - 50°C = 30°C, c = 4.18 J/g/K

Energy required to raise 92426035.2 g water by 30° C = 92426035.2 × 4.18 × 30

Energy required = 1.16 × 10¹⁰ J

Hence, number of moles of photons required = 1.16 × 10¹⁰ J/2.30 × 10⁵ J/mol

Number of moles of photons required = 5.04 × 10⁴ moles

5 0
3 years ago
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