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german
3 years ago
14

If r⃗ =bt2i^+ct3j^, where b and c are positive constants, when does the velocity vector make an angle of 45.0∘ with the x- and y

-axes?
Chemistry
1 answer:
Papessa [141]3 years ago
5 0
The displacement function is given by : r = bt2i + ct3j meters
The velocity function is the derivative of the displacement:
v = r' = 2bti + 3ct2j meter / second

Now, for <span> the velocity vector make an angle of 45.0∘ with the x- and y-axes, the i and j components have to be equal:
</span>2bt = 3ct2 (Now solve for t)
2bt = t (3ct)
either t = zero (rejected)
or t = 2b / 3c seconds (accepted)
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Calculate the number of moles of each compound, given the number of molecules.
brilliants [131]

Answer:

1.  2.46\times 10^{21} molecules of CO₂

2. 10⁴ molecules of H₂O

3. 8.75×10³² molecules of C₆H₁₂O₆

Explanation:

1.  2.46\times 10^{21} molecules of CO₂

2. 10⁴ molecules of H₂O

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Which equation correctly relates kinetic energy mass and velocity
jolli1 [7]

Answer: The equation for kinetic energy is \frac{1}{2}mv^2

Explanation:

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Mathematically,

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3 0
3 years ago
Read 2 more answers
What is the concentration of hydronium ion in a 0.121 M HCl solution?
kogti [31]

Answer:

C) 0.121 M

Explanation:

HCl + H₂O = H₃O⁺ + OH⁻

.121M              .121M

HCl is a strong acid . It will dissociate almost 100 % so the concentration of acid and hydronium ion formed will be equal . It is to be noted that hydronium ion is formed due to association of H⁺ and H₂O . H⁺ is formed due to ionisation of HCl .

So concentrtion of hydronium ion ( H₃O⁺ ) will be .121 M.

7 0
3 years ago
In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

7 0
3 years ago
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