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ira [324]
3 years ago
15

Circle the letter of each expression that has four significant figures. a. 1.25 104 b. 12.51

Chemistry
1 answer:
DiKsa [7]3 years ago
4 0

B is the answer

1 2 5 1

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Please help me with this.
zlopas [31]

Answer A: Connect a wire coil to an ammeter. Move a bar magnet into and out of the wire coil as you observe the ammeter.

6 0
2 years ago
If a sample containing 18.1 g of NH3 is reacted with 90.4 g of
USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃  = 18.1g

Mass of Cu₂O  = 90.4g

Unknown:

Limiting reactant  = ?

Mass of N₂ formed  = ?

Solution:

The reaction equation is given as:

       Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

  Number of moles = \frac{mass}{molar mass}  

Molar mass of Cu₂O = 2(63.6) + 16  = 143.2g/mol

Molar mass of NH₃  = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = \frac{18.1}{143.2}   = 0.13moles

Number of moles of NH₃   = \frac{90.4}{17}   = 5.32moles

  From this reaction;

       1 mole of  Cu₂O combines with 2 mole of NH₃

So   0.13moles of  Cu₂O will combine with 0.13 x 2 mole of NH₃

                                              = 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

   Mass = number of moles x molar mass

    1 mole of Cu₂O  will produce 1 mole of N₂

    0.13 mole of Cu₂O  will produce 0.13 mole of N₂

    Mass  = 0.13 x (2 x 14) = 3.64g

5 0
3 years ago
Which two compounds are classified as bases by the Brønsted-Lowry definition, but not by the Arrhenius definition, and why?
konstantin123 [22]

Answer: Ammonia (NH3) and sodium carbonate (Na2CO3), because they accept hydrogen ions but lack hydroxide ions.

Explanation:

i took the test and got it correct :) hope this helps

6 0
3 years ago
GIVING OUT BRAINLIEST
d1i1m1o1n [39]

Answer:

it increases and is perpendicular to the motion of the wave.

4 0
3 years ago
Read 2 more answers
2NaOH+CO2----->Na2CO3+H2O
kondaur [170]

Answer:

13.2 g Na2CO3

Explanation:

Convert 10.0 g NaOH to mol.

10.0 g x 1 mol/39.997 g = 0.250 mol

Use mol ration given by the equation: 2 mol NaOH to 1 mol Na2CO3

0.250 mol NaOH x 1 mol Na2CO3/2 mol NaOH = 0.125 mol Na2CO3

Finally, convert the moles of Na2CO3 to grams.

0.125 mol Na2CO3 x 105.99 g/1 mol = 13.2 g

8 0
3 years ago
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