Explanation:
<em>2</em><em>1</em><em> </em><em>piec</em><em>e</em><em> </em><em>=</em><em>4</em><em>5</em><em> </em><em>mins</em>
<em>100</em><em> </em><em>pieces</em><em>=</em><em> </em><em>x</em>
<em>on</em><em> </em><em>cross</em><em> </em><em>mul</em><em>tiplicati</em><em>on</em>
<em>2</em><em>1</em><em>x</em><em>=</em><em>4</em><em>5</em><em>×</em><em>1</em><em>0</em><em>0</em><em>m</em><em>i</em><em>n</em><em>s</em>
<em>2</em><em>1</em><em>x</em><em> </em><em>=</em><em>4</em><em>5</em><em>0</em><em>0</em><em>m</em><em>i</em><em>n</em><em>s</em>
<em>x</em><em> </em><em>=</em><em>4</em><em>5</em><em>0</em><em>0</em><em>m</em><em>i</em><em>n</em><em>s</em><em>÷</em><em>2</em><em>1</em>
<em> </em><em> </em><em>=</em><em>2</em><em>1</em><em>4</em><em>.</em><em>3</em><em>m</em><em>i</em><em>n</em><em>s</em><em> </em><em>or</em><em> </em><em> </em><em>3</em><em>.</em><em>5</em><em>7</em><em>h</em><em>r</em><em>s</em>
They move because of convection currents in the mantle
Answer:
The answer is: 11759 Hz
Explanation:
Given: Chemical shift: δ = 211.5 ppm, Spectrometer frequency = 556 MHz = 556 × 10⁶ Hz
In NMR spectroscopy, the chemical shift (δ), expressed in ppm, of a given nucleus is given by the equation:
<u>Therefore, the signal is at 11759 Hz from the TMS.</u>
Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
Here we have explain that the maximum possible electrons present in nitrogen valence shell is 8 whereas in phosphorous 12 valence electrons are present.
Although both nitrogen (N) and phosphorous (P) belongs to the same series there are several properties which are different between both the element. The number of electrons present in nitrogen is seven which are present in the -s and -p orbitals. The electronic configuration of nitrogen is 1s²2s²2p³. In which the outermost electrons are the valence electrons i.e. 5 valence electrons are present. The maximum orbitals are possible under the principal quantum number 2 are -s and -p orbitals. Now the maximum capacity of the p orbital to contain 6 electrons, as it is half filled in nitrogen another 3 electrons can be incorporated. Thus the maximum number of electrons can be present in nitrogen is 10 among which 8 is the valence electrons.
On the other hand there are 15 electrons in phosphorous the electronic configuration is 1s²2s²2p⁶3s²3p³. Now the principal quantum number 3 can have three orbitals -s, -p and -d. So another 13 electrons can be incorporated (3 in -p orbital and 10 in -d orbital) among which upto 12 electrons can be its valence electrons.