Answer:
5 electron groups, see saw
Explanation:
During the formation of SF4, the sulfur atom usually bonds with each of four fluorine atoms where 8 of valence electrons are used. The four fluorine atoms have 3 lone pairs of electrons in its octet which will further utilize 24 valence electrons. In addition, two electrons are present as a lone pair on the sulfur atom. We can determine sulfur’s hybridization state by counting of the number of regions of electron density on sulphur (the central atom in the molecule). When bonding takes place there is a formation of 4 single bonds to sulfur and it has 1 lone pair. Looking at this, we can say that the number of regions of electron density is 5. The hybridization state is sp3d.
SF4 molecular geometry is seesaw with one pair of valence electrons. The molecule is polar. The equatorial fluorine atoms have 102° bond angles instead of the actual 120° angle. The axial fluorine atom angle is 173° instead of the actual 180° bond angle.
Answer: B. An experiment that directly tests the hypothesis
Explanation:
Yasir didn’t really test his experiment he only ask people for the opinion on which color they like so he forget to actually do the experiment therefore the answer is b.
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Answer:
16.02 g
Explanation:
the balanced equation for the decomposition of CuCO₃ is as follows
CuCO₃ --> CuO + CO₂
molar ratio of CuCO₃ to CO₂ is 1:1
number of CuCO₃ moles decomposed - 45 g / 123.5 g/mol = 0.364 mol
according to the molar ratio
1 mol of CuCO₃ decomposes to form 1 mol of CO₂
therefore 0.364 mol of CuCO₃ decomposes to form 0.364 mol of CO₂
number of CO₂ moles produced - 0.364 mol
therefore mass of CO₂ produced - 0.364 mol x 44 g/mol = 16.02 g
16.02 g of CO₂ produced
Answer:
6.48 L
Explanation:
From the question,
Applying
PV/T = P'V'/T'......................... Equation 1
P = initial pressure of the helium balloon, V = Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.
make V' the subject of the equation
V' = PVT'/P'T......................... Equation 2
Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm
Substitute these values into equation 2
V' = (4.5×1×253)/(0.6×293)
V' = 1138.5/175.8
V' = 6.48 L