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Gnom [1K]
3 years ago
15

a rope breaks when the tension reaches 305N what is the largest largest mass it can swing at 14.2m/s in a circle of radius 0.425

m?(unit=kg)
Chemistry
1 answer:
Iteru [2.4K]3 years ago
6 0

Answer:

0.643 kg

Explanation:

Apply Newton's second law to the mass:

∑F = ma

T = m v² / r

305 N = m (14.2 m/s)² / 0.425 m

m = 0.643 kg

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the answer is 360 g H2O with a 90.3 yeild percentage

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Under what conditions of temperature and pressure is a gas least soluble in water
snow_tiger [21]

The conditions of temperature and pressure in which a gas least soluble in water is low pressure and high temperature.

<h3>What is Henry Law?</h3>

The amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid, according to Henry's law.

From this law it is clear that:

  • As the pressure of the gas increases solubility of the gas on the liquid also increases.

But if the temperature of the liquid decreases then the solubility of the gas also increases.

Hence at low pressure and high temperature, gas is least soluble.

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2 years ago
In which set do all elements tend to form anions in binary ionic compounds? li, na, k n, o, i c, s, pb k, fe, br?
pav-90 [236]

Atoms or molecule after gaining of electron possesses negative charge and is known as anion.

For the given sets:

  • Li, Na, K

The given elements are alkali metals and have tendency to lose electrons easily and form cations.

  • N, O, I

The given elements are non-metals and are electronegative. So, they gain electrons easily and form anion.

  • C, S, Pb

Carbon has tendency to form bond by sharing of electrons, Sulfur has tendency to gain electrons and form anion whereas Lead has tendency to lose electron.

  • K, Fe, Br

Potassium and Iron has tendency to lose electron and form cation whereas Bromine has tendency to gain electron to form anion.

Hence, from the given sets, all elements of set: N, O, I have tendency to form anions in binary ionic compounds.

8 0
3 years ago
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combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

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What type of atmospheric conditions does climate describe?
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Explanation  believe the answer is A. Average.

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