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Tanya [424]
3 years ago
5

Part B Now that we have put a coefficient of 3 in front of CaSO4, what coefficient should go in front of CaCl2 to balance calciu

m (Ca)? 3CaSO4+AlCl3→?CaCl2+Al2(SO4)3 Express your answer numerically as an integer.
Chemistry
1 answer:
Jet001 [13]3 years ago
4 0

Answer: The coefficient must be 3.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

CaSO_4+AlCl_3\rightarrow CaCl_2+Al_2(SO_4)_3

The number of atoms of each element has to be same on both sides, thus the balanced chemical reaction will be:

3CaSO_4+2AlCl_3\rightarrow 3CaCl_2+Al_2(SO_4)_3

Thus a coefficient of 3 should go in front of CaCl_2 to balance calcium (Ca).

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1.025 moles of NaCl are present in a solution with a molarity of 8.20M and 125 mL of solution. Details about molarity can be found below.

<h3>How to calculate molarity?</h3>

The number of moles of a solution can be calculated by multiplying the molarity of the solution by its volume. That is;

no of moles = molarity × volume

According to this question, a solution has a molarity of 8.20M and volume of 125 mL of solution.

no of moles = 8.20 × 0.125

no of moles = 1.025moles

Therefore, 1.025 moles of NaCl are present in a solution with a molarity of 8.20M and 125 mL of solution.

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2 years ago
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MrRissso [65]

Answer:   17 (a) 360,000 ms         18 (a) 0.245 s

                     (b) 4.800 kg                   (b) 500 cm

                     (c) 560.0 dm                  (c) 68.00 m

                     (d) 72,000 mg                (d) 0.025 Mg

<u>Explanation:</u>

Use the following table.  Note the direction you are going to:

Right means move decimal to the right. Left means move decimal to the left.

               \boxed{\begin{array}{c|c|c|c|c|c|c|c|c|c}Mega&x&x&kilo&x&deka&unit&deci&centi&milli\\M\_\_&&&k\_\_&&dk\_\_&\_\_\_&d\_\_\)&c\_\_&m\_\_\end{array}}

**********************************************************************************************

17(a) 360 s to ms                                                        o       →         →        →o

                            \boxed{\begin{array}{c|c|c|c|c|c|c|c|c|c}Mega&x&x&kilo&x&deka&unit&deci&centi&milli\\M\_\_&&&k\_\_&&dk\_\_&\_\_\_&d\_\_\)&c\_\_&m\_\_\end{array}}

Move the decimal 3 places to the right   -->   360_ _ _.

Fill in the blanks with zeroes   --> 360,000

  (b) 4800 g to kg                            o←    ←     ←         o

                             \boxed{\begin{array}{c|c|c|c|c|c|c|c|c|c}Mega&x&x&kilo&x&deka&unit&deci&centi&milli\\M\_\_&&&k\_\_&&dk\_\_&\_\_\_&d\_\_\)&c\_\_&m\_\_\end{array}}

Move the decimal 3 places to the left   --> 4.<u>8</u> <u>0</u> <u>0</u>

<u />

 (c) 5600 dm to m                                                       o←     o

                             \boxed{\begin{array}{c|c|c|c|c|c|c|c|c|c}Mega&x&x&kilo&x&deka&unit&deci&centi&milli\\M\_\_&&&k\_\_&&dk\_\_&\_\_\_&d\_\_\)&c\_\_&m\_\_\end{array}}

Move the decimal 1 place to the left   --> 560. <u>0</u>

<u />

 (d)  72 g to mg                                                         o         →        →        →o

                            \boxed{\begin{array}{c|c|c|c|c|c|c|c|c|c}Mega&x&x&kilo&x&deka&unit&deci&centi&milli\\M\_\_&&&k\_\_&&dk\_\_&\_\_\_&d\_\_\)&c\_\_&m\_\_\end{array}}

Move the decimal 3 places to the right   --> 72 _ _ _.

Fill in the blanks with zeroes   -->     72,000

***********************************************************************************************

18 (a) 245 ms to s                                                       o←       ←       ←        o

                             \boxed{\begin{array}{c|c|c|c|c|c|c|c|c|c}Mega&x&x&kilo&x&deka&unit&deci&centi&milli\\M\_\_&&&k\_\_&&dk\_\_&\_\_\_&d\_\_\)&c\_\_&m\_\_\end{array}}

Move the decimal 3 places to the left -->   . <u>2</u> <u>4</u> <u>5</u>

     (b) 5 m to cm                                                        o        →        →o

                             \boxed{\begin{array}{c|c|c|c|c|c|c|c|c|c}Mega&x&x&kilo&x&deka&unit&deci&centi&milli\\M\_\_&&&k\_\_&&dk\_\_&\_\_\_&d\_\_\)&c\_\_&m\_\_\end{array}}

Move the decimal 2 places to the right   -->   5 _ _.

Fill in the blanks with zeroes   -->    500

      (c) 6800 cm to m                                                 o←      ←        o

                             \boxed{\begin{array}{c|c|c|c|c|c|c|c|c|c}Mega&x&x&kilo&x&deka&unit&deci&centi&milli\\M\_\_&&&k\_\_&&dk\_\_&\_\_\_&d\_\_\)&c\_\_&m\_\_\end{array}}

Move the decimal 2 places to the left   -->   68. <u>0</u> <u>0 </u>

<u />

    (d)  25 kg to Mg    o←    ←    ←    o

                            \boxed{\begin{array}{c|c|c|c|c|c|c|c|c|c}Mega&x&x&kilo&x&deka&unit&deci&centi&milli\\M\_\_&&&k\_\_&&dk\_\_&\_\_\_&d\_\_\)&c\_\_&m\_\_\end{array}}

Move the decimal 3 places to the left -->   . _ 25

Fill in the blanks with zeroes  -->   .025

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4 years ago
Why is vsepr theory not needed to predict the shape of hcl?
NeX [460]

<span>Two-dimensional Lewis dot formulas help us understand the bonding within a molecule or polyatomic ion, but they do not give us a sense of the 3-dimentional shape of the particle. Valence Shell Electron Repulsion Theory (VSEPR) is often used to predict particle shape from a Lewis dot formula.The VSEPR theory focuses on the idea that electron pairs and electrons repel one another and that these repulsions are smallest when the electron pairs or groups of electron pairs are as far apart as possible. This will then be the most stable form or shape of a molecule or ion.We know from a study of Lewis formulas that molecules and polyatomic ions may contain single bonds, double bonds, triple bonds, and "lone pairs" of electrons that are not used for bonding. We also know that a particle contains one or more "central atoms" around which the rest of the atoms are arranged; we know that the rest of the atoms are bonded either directly or through other atoms to this center atom.In the VSEPR theory approach to particle shapes, you focus on two things.<span><span><span>the </span>central atom</span><span><span>the </span>number of different electron groups<span> around the central atom</span></span></span>The arrangement in space (geometry ) of the electron groups around a center atom controls the overall shape of a particle because all bonds radiate out from the central atom of the particle.<span>An electron group may be 1 pair of electrons (single bond or lone pair), 2 pairs (double bond) or 3 pairs (triple bond). The carbonate ion, for example, has one double bond and two single bonds attached to the center carbon atom. Thus, there are </span>3 groups<span> of electrons around the C even though there are 4 pairs of electrons on carbon. Two pairs of electrons point in the same direction, the double bond to O. The other two pairs go in two other directions, one pair to each remaining O. One double bond and two single bonds on the center atom are considered to be 3 electron groups.</span><span> </span></span>
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