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GrogVix [38]
3 years ago
15

A commercial airplane typically flies at a height of 37,000 feet. If an average cell phone was dropped from an airplane, about h

ow many seconds would it take to reach the ground? Round to the nearest whole number.
Chemistry
2 answers:
saul85 [17]3 years ago
6 0

Answer:

its b i just did it .

sveticcg [70]3 years ago
5 0
Well depends on how much the phone weighs but I’m sure it’s going off 1 lb so 1000 feet a second so your answer should be 37 seconds
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Three isotopes of sulfur are sulfur-32, sulfur-33, and sulfur-34. write the complete symbol for each isotope, including the atom
Maurinko [17]
Sulfur is an element in the periodic table that has a chemical symbol of S. This element is the 16th element in periodic table. This means that the atomic number or number of protons in the nucleus of the atom is equal to 16.

The number following the name of the element is the mass number. The following are the complete symbol that are arranged as follows:
 chemical symbol - atomic number - mass number

*Sulfur-32
         S - 16 - 32

*Sulfur-33
        S - 16 - 33

*Sulfur-34
        S - 16 - 34
8 0
3 years ago
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Does Mitosis have body cells or sex cells?
Rufina [12.5K]
2 identical daughter cells
5 0
4 years ago
1.what happens in the nucleus of an atom when an alpha particle is emitted?
coldgirl [10]
1) The nucleus of an atom loses 2 protons and 4 neutrons.
2) The nucleus of an atom gains a proton and it's neutrons remain the same.
7 0
3 years ago
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What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 giv
statuscvo [17]

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

7 0
3 years ago
Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4
inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

M=\frac{n_{solute}}{V_{solution}}

So you can solve for the moles of the solute:

n_{solute}=M*V_{solution}

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

n_{solute}=1mol/L*1.5L=1.5mol

But this is just a supposition.

Regards.

4 0
3 years ago
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