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Ne4ueva [31]
3 years ago
15

During active transport, molecules move from areas of ______________ to _______________. This requires the use of _________.

Chemistry
1 answer:
MissTica3 years ago
8 0
Molecules move from areas of low concentration to areas of high concentration. This requires the use of energy.
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150.0 mL of a gas is collected at 50 kPa, What will the volume be at 125kpa?
slava [35]

Answer:

C. 60 mL

Explanation:

Boyle's Law P₁V₁ = P₂V₂

                   (50kPa)(150.0mL) = (125kPa)(V₂)

                   V₂ = 7500/125

                   V₂ = 60 mL

4 0
2 years ago
What is the value for the potential energyfor a n = 6 Bohr orbit electron in Joules?
ioda

Answer:

Potential energy for n = 6 Bohr orbit electron is -1.21*10⁻¹⁹J

Explanation:

As per the Bohr model, the potential energy of electron in an nth orbit is given as:

PE_{n} = -\frac{kZe^{2}}{r_{n}}

here:

k = Coulomb's constant = 9*10⁹ Nm2/C2

Z = nuclear charge

e = electron charge = 1.6*10⁻¹⁹ C

r(n) = radius of the nth orbit = n²(5.29*10⁻¹¹m)

Substituting for k, Z(= 1), e and r(n) in the above equation gives:

PE_{6} = -\frac{9*10^{9}Nm2/C2*1*(1.6*10^{-19}C)^{2}}{(6)^{2}*5.29*10^{-11}m}=-1.21*10^{-19}J

3 0
2 years ago
By examining electron domain geometry, one can determine that the ammonia molecule (NH3) has
Annette [7]

has only one sp2 orbitals, 

<span>Only one sp2 shell is filled up because it ammonia has only 5 valence electrons. Considering ammonia's trigonometrical pyramid shape, 3 hydrogen atoms are attached to Nitrogen. A lone pair remains unattached to any other atom and results to hybridization of sp3 orbitals.</span>

8 0
3 years ago
a mixture is made by combining 1.64 lb of salt and 4.66 lb of water. what is the percentage of salt (by mass) in this mixture?
const2013 [10]
1.64 + 4.66 = 6.3
1.64/6.3 = 0.2603174603 = 26% 
5 0
3 years ago
Read 2 more answers
At 25 oC, hydrogen iodide breaks down very slowly to hydrogen gas and iodine vapor with a rate constant of 2.4 x 10-21L/mol.s. I
Ratling [72]

Answer:

\large \boxed{4.6 \times 10^{21}\text{ s}}

Explanation:

Whenever a question asks you, "What is the concentration after a given time?" or something like that, you must use the appropriate integrated rate law expression.

The reaction is 2nd order, because the units of k are L·mol⁻¹s⁻¹.

The integrated rate law for a second-order reaction is  

\dfrac{1}{\text{[A]}} =\dfrac{1}{\text{[A]}_{0}}+ kt

Data:

   k = 2.4 × 10⁻²¹ L·mol⁻¹s⁻¹

[A]₀ = 0.0100     mol·L⁻¹

[A] = 0.009 00 mol·L⁻¹

Calculation :

\begin{array}{rcl}\dfrac{1}{\text{[A]}} & = & \dfrac{1}{\text{[A]}_{0}}+ kt\\\\\dfrac{1}{0.00900 }& = & \dfrac{1}{0.0100} + 2.4 \times 10^{-21} \, t\\\\111.1&=& 100.0 + 2.4 \times 10^{-21} \, t\\\\11.1& = & 2.4 \times 10^{-21} \, t\\t & = & \dfrac{11.1}{ 2.4 \times 10^{-21}}\\\\& = & \mathbf{4.6 \times 10^{21}}\textbf{ s}\\\end{array}\\\text{It will take $\large \boxed{\mathbf{4.6 \times 10^{21}}\textbf{ s}}$ for the HI to decompose}

8 0
3 years ago
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