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kondaur [170]
3 years ago
8

How many moles of gas would be present in a gas trapped within a 100.0 mL vessel at 25.0°C at a pressure of 2.50 atmospheres?

Chemistry
1 answer:
nika2105 [10]3 years ago
7 0

Answer:

bjbchgchhvvjnkvgx

Explanation:

njvg id xgg

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What is the main difference between a homogeneous mixture and a heterogeneous mixture?
Sholpan [36]
The answer is <span>The components of a homogeneous mixture are evenly distributed.

In a homogeneous mixture, all components are evenly distributed. They are known as solutions. In a heterogeneous mixture, components are not evenly distributed. It consists of visibly different components. For example, milk is the homogeneous mixture, you cannot see its particles. But milk and cereals are the heterogeneous mixtures.</span>
7 0
3 years ago
16. If the velocity of hydrogen molecule is 5 x 10^4cm sec-¹, then its de Broglie wavelength is
Aleks04 [339]

Answer:

Correct option is B)

According to de-Broglie, 

λ=mvh=6.023×10232×5×104cm/sec6.62×10−27ergsec=4×10−8cm=4Ao

5 0
2 years ago
The following choices contain densities of various materials. Based on these data, which will yield the fastest seismic velocity
Eduardwww [97]

Answer:

D. 5.6 g/cm^3

Explanation:

On the average seismic velocity increases with increase in depth due higher the pressure and more compaction

sand and shales in the Niger Delta Basin density–velocity relationship is

P = 0.31×V^0.25

A derivation of the original Gardner equation to calculate the average densities for sands and shales in wells.

ρ = α ×V^β

where

ρ = bulk density in g/cm3,

V = P-wave velocity,

α = 0.31 for V (m/s) and 0.23 for V(ft/s) and

β = 0.25.

Such that

ρ = 0.31 ×V^0.25

So the fastest seismic velocity will be in the densest material which is D. 5.6 g/cm3

5 0
3 years ago
Alkenes ______
Vlada [557]

Answer:

B

Explanation:

alkene is no polar compounds

and insoluble in water

6 0
3 years ago
20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the
kirill115 [55]

Answer:

Choice A: approximately 0.12\; \rm M.

Explanation:

Note that the unit of concentration, \rm M, typically refers to moles per liter (that is: 1\; \rm M = 1\; \rm mol\cdot L^{-1}.)

On the other hand, the volume of the two solutions in this question are apparently given in \rm cm^3, which is the same as \rm mL (that is: 1\; \rm cm^{3} = 1\; \rm mL.) Convert the unit of volume to liters:

  • V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L.
  • V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L.

Calculate the number of moles of \rm H_2SO_4 formula units in that 0.02\; \rm L of the 0.09\; \rm M solution:

\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}.

Note that \rm H_2SO_4 (sulfuric acid) is a diprotic acid. When one mole of \rm H_2SO_4 completely dissolves in water, two moles of \rm H^{+} ions will be released.

On the other hand, \rm NaOH (sodium hydroxide) is a monoprotic base. When one mole of \rm NaOH formula units completely dissolve in water, only one mole of \rm OH^{-} ions will be released.

\rm H^{+} ions and \rm OH^{-} ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid \rm H_2SO_4 dissolves in water completely, it will take two moles of \rm OH^{-} to neutralize that two moles of \rm H^{+} produced. On the other hand, two moles formula units of the monoprotic base \rm NaOH will be required to produce that two moles of \rm OH^{-}. Therefore, \rm NaOH and \rm H_2SO_4 formula units would neutralize each other at a two-to-one ratio.

\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O.

\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2.

Previous calculations show that 0.0018\; \rm mol of \rm H_2SO_4 was produced. Calculate the number of moles of \rm NaOH formula units required to neutralize that

\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}.

Calculate the concentration of a 0.03\; \rm L solution that contains exactly 0.0036\; \rm mol of \rm NaOH formula units:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}.

3 0
3 years ago
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