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Aloiza [94]
2 years ago
15

The number of protons plus the number of neutrons equal the ______

Chemistry
2 answers:
NARA [144]2 years ago
7 0
the answer is: Atomic number
Serga [27]2 years ago
4 0
Atomic weight I think this is right
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a.endothermic reaction

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A plane on the ground will weigh
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Element X is in group 2, and element Y is in group 7, of the periodic table. which ions will be present in the compound formed w
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8 0
2 years ago
5. A penny weighs about 2.5 g. How many moles of pennies would be required to equal the mass of the moon (7.3x10^24 kg)
schepotkina [342]

5. 1.16 x 10^{26} moles  moles of pennies would be required to equal the mass of the moon.

6. 12.86 moles of ethanol are in a 750 ml bottle of vodka.

Explanation:

5 .Data given:

mass of penny = 2.5 grams

atomic mass of penny = 62.93 grams/mole

moles present in mass of the moon given as = 7.3 x 10^{24} kg

number of moles = \frac{mass}{atomic mass of  1 mole}

number of moles = \frac{2.5}{62.93}

0.039 moles of penny is present in 2.5 grams

0.039 moles of penny in 2.5 grams of it

so, x moles in 7.3 X 10^{27}  grams

\frac{0.039}{2.5} =\frac{x}{7.3 X 10^{27} }

x = 1.16 x 10^{26} moles

so when the mass of the penny given is equal to the mass of moon, number of moles of penny present is 1.1 x 10^{26} .

6.

Given:

vodka = 40% ethanol

volume of vodka bottle = 750 ml

moles of ethanol =?

density of ethanol =0.79 g/ml

atomic mass of ethanol = 46.07 grams/mole

so, from the density of ethanol given we can calculate how much ethanol is present in the solution.

density = \frac{mass}{volume}

density x volume = mass

0.79 x 750 = 592.5 grams

number of moles = \frac{mass}{atomic mass of 1 mole}

number of moles of ethanol = \frac{592.5}{46.07}

                                               = 12.86 moles of ethanol

6 0
2 years ago
Could someone help pls ​
drek231 [11]
<h3>Solution-:</h3>

  • option D
  • maintains a constant volume.

#<em>o</em><em>f</em><em>f</em><em>i</em><em>c</em><em>a</em><em>i</em><em>l</em><em> </em><em>Nazo</em>

<em>ll </em><em>Radhe</em><em> Radhe</em><em> ll</em>

4 0
2 years ago
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