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3 years ago

A mass of the crucible (g) 88.000g

1 answer:

4 0

A)5.000g
b)mole=mass/molarmass
   molar mass of H2O=18g
   mass of water=5g
b)mole of water=5.000/18.000=0.2778g
c)mass of salt=2.342g
d)mole of salt=mass/molar mass
   pure salt is NaCl
   molar mass of NaCl=58.50g
d)mole of salt=2.342/58.5=0.040mol
e)mole ratio=0.27:0.04

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For the following reaction, identify the element that was oxidized, the element that was reduced, and the reducing agent. Give a

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3 years ago

A substance with a high [h ] would likely have which additional characteristics?

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Answer:

Is [h] really [H]? What characteristics are mentioned?

Explanation:

A high [H} means high acid concentration.

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2 years ago

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3 years ago

Read 2 more answers

calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?

inessss [21]

Answer:

$\Delta E=E_{final}-E_{initial}$

$\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}$

$\Delta E=-1312[0.111-1]KJ mol^{-1}$

$\Delta E=1166 KJ mol^{-1}$

$\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}$

$=193623 \times 10^{-23} \frac {J}{photon}$

$\Delta E=1.93623 \times 10^{-18} \frac {J}{photon}$

$\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}$

h is planck's constant

c is the speed of light

λ is the wavelength of light

$\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18} J/photon)}$

Wavelength

$\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m} =103 nm (Answer)$

Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is 103 nm.

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3 years ago

()

const2013 [10]

Answer:

$\large \boxed{\text{-1276 kJ/mol}}$

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

CH₃CH₂OH + 3O₂ ⟶ 2CO₂ + 3H₂O

Bonds: 5C-H 1C-C 1C-O 1O-H 3O=O 4C=O 6O-H

D/kJ·mol⁻¹: 413 347 358 467 495 799 467

$\Delta H = \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\\sum{D_{\text{reactants}}} = 5 \times 413 + 1 \times 347 + 1 \times 358 + 1 \times 467 + 3 \times 495 = 3237 + 1485\\=\text{4722 kJ}\\\sum{D_{\text{products}}} = 4 \times 799 + 6 \times 467 =3196 + 2802 = \text{5998 kJ}\\\Delta H = 4722 - 5998= \textbf{-1276 kJ} \\ \text{The overall energy change is $\large \boxed{\textbf{-1276 kJ/mol}}$}.$

6 0

3 years ago