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AfilCa [17]
3 years ago
11

Please help me match these to the boxes

Chemistry
1 answer:
mamaluj [8]3 years ago
6 0

Answer:

answers from left to right:

decomposition,decomposition,synthesis,replacement,synthesis,replacement

Explanation:

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A Balanced Equation: The total amount of atoms involved in the chemical reaction changes depending on the reaction. True or fals
Stells [14]
False is the answer



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5 0
3 years ago
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N2H4 + N2O4 --> N2 + H2O
ahrayia [7]

Answer:

  1. The limiting reagent is N2O4
  2. 14,09g

Explanation:

  • First, we adjust the reaction.

2N_{2} H_{4} + N_{2} O_{4} ⇄6N_{2} +  4H_{2}O

  • Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using N_{2} H_{4} to form H_{2}O

               molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol}  H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4}  }{32,04\frac{g}{mol}  N_{2} H_{4} }

                                           molH_{2} O = 1, 125 mol

Using N_{2} O_{4} to form H_{2} O

              molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol}  H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4}  }{92\frac{g}{mol}  N_{2} O_{4} }

                                           molH_{2} O = 0,783 mol

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴                          MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}

                                      MassOfH_{2}O = 14,09g

5 0
3 years ago
No work is done when an energy transformation takes place. <br><br><br> True or False
Finger [1]
I would go with the statement is : False

-Hope this helps.

6 0
3 years ago
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Your veterinarian is administering a sedative to your 50 pound dog. The sedative is mixed in saline solution. Unfortunately the
Olin [163]

Answer:

26.25 mL

Explanation:

This is a dilution problem. First, let us calculate the volume of final solution needed:

The dog weighs 50 pounds and the sedative is administered at 0/7 ml per pound. Hence:

50 x 0.7 = 35 mL

A total volume of 35 mL, 2.5% solution of the sedative will be needed.

But 10% solution is available. There needs to be a dilution with saline water, but what volume of the 10% solution would be diluted?

initial volume = ?

final volume = 35 mL

initial concentration = 10%

final concentration = 2.5%

Using dilution equation:

initial concentration x initial volume = final concentration x final volume

initial volume = \frac{final concentration*final volume}{initial concentration}

                     = 2.5 x 35/10 = 8.75 mL

Hence, 8.75 mL of the 10% pre-mixed sedative will be required.

But 35 mL is needed? The 8.75 mL is marked up to 35 mL with saline water.

35 - 8.75 = 26.25 mL

<em>Therefore, 26.25 mL of saline water will be added to 8.75 mL of the 10% pre-mixed sedative to give 2.5%, 35 mL needed for the dog.</em>

5 0
2 years ago
The last column of the periodic table contains the noble gases elements that do not easily form chemical bonds. Why don’t these
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Answer:

its is because the chemicals are very diffraction from each other so which makes it difficult to form a chemical bond.

6 0
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