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iren2701 [21]
3 years ago
7

What type of bond is H2Se

Chemistry
1 answer:
Vinvika [58]3 years ago
6 0

Answer:The lone pair of electrons takes up more space than a regular bonding pair since it it is not confined to be between two atoms, so it adds coulombic repulsion to the bonding pairs and compresses the angle. Therefore, the bond angle is less than the standard 109.5∘ . It is actually 97.7∘

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flourine

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Form the complementary solution to the homogeneous equation. yc(t) = Cl [ ] +C2 Construct the (simplest) particular olution of t
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Explain how creativity can play a role in the construction of scientific questions and hypotheses.
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3 years ago
What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
lyudmila [28]

Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

4 0
3 years ago
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