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QveST [7]
2 years ago
11

Determine the mass of the following samples (remember mass => grams)

Chemistry
1 answer:
kakasveta [241]2 years ago
5 0

Mass is the amount of content available in the matter. The mass of the moles of samples given are

2 moles Co = 117.86 g Co

5 moles H2 =  10.10 g  H₂

2.1 moles H₂O = 37.84 g  H₂O

<h3>What is a mole?</h3>

A  mole can be atoms or molecules or ions or any substance.

The mass of one mole of Co is 58.93 g.

Then, 2 moles of Co will be 2 x 58.93 g = 117.86 g Co

The mass of one mole of H₂ is 2.02 g.

Then, 5 moles of  H₂ will be 5 x 2.02 g =  10.10 g  H₂

The mass of one mole of H₂O is 18.02 g .

Then, 2.1 moles of  H₂O will be 2.1 x 18.02 g =  37.84 g  H₂O

Thus, the mass of the following samples in grams

2 moles Co = 117.86 g Co

5 moles  H₂=  10.10 g  H₂

2.1 moles H₂O = 37.84 g  H₂O

Learn more about mole.

brainly.com/question/26416088

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MariettaO [177]

 The atomic number increases moving left to right across a period and subsequently so does the effective nuclear charge. Therefore, moving left to right across a period the nucleus has a greater pull on the outer electrons and the atomic radii decreases.

7 0
3 years ago
According to the phase diagram for H₂O, what happens to the phases of
sergejj [24]

Answer:

"A", "water changes from a gas to a solid to a liquid", according to this phase diagram, at at 0°C, as pressure is increased from 0atm to 10atm.

Explanation:

The question asks what happens at 0°C, as pressure is increased from 0atm to 10atm.

According to the question, the temperature is held constant.  The pressure changes.  In the phase diagram, we find the temperature 0°C on the horizontal axis, and all points where the temperature are 0°C are along that vertical line.

Since the pressure starts at 0atm and increases to 10atm, we start at the bottom, and move upward along that line, to see what phases of matter the substance changes to.

At the bottom, it is initially in a "gas" phase.  As it moves up, it transitions to a "solid" phase.  Later, as it continues moving up, it changes again into a "liquid" phase.

Thus, the answer would be "A", "water changes from a gas to a solid to a liquid", according to this phase diagram, at at 0°C, as pressure is increased from 0atm to 10atm.

8 0
2 years ago
What is the cell potential of an electrochemical cell that has the half-reactions shown below?
Karo-lina-s [1.5K]

Answer:

E°(Ag⁺/Fe°) = 0.836 volt

Explanation:

3Ag⁺ + 3e⁻ => Ag°;            E° = +0.800 volt

Fe° => Fe⁺³ + 3e⁻ ;             E° = -0.036 volt

_________________________________

Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...    

E°(Ag⁺/Fe°) = E°(Ag⁺) - E°(Fe°) = 0.800v - ( -0.036v) = 0.836 volt

4 0
3 years ago
The reaction is found to be first order in IO3−, first order in SO32−, and first order in H+. If [IO3−]=x, [SO32−]=y, and [H+]=z
ollegr [7]

Answer:

rate = kxyz

Explanation:

It is worth knowing that the rate low can only be determined by experimentation only not by just balancing equations. So here we are told that all the reactants x , y and z are all first order. This is important because we use this as exponents. That is why the exponents of all the reactants will be 1.

rate = kxyz

7 0
3 years ago
Escriba en termino de moles, de moléculas y de masa las siguientes ecuaciones a. Fe +2HCl ________ FeCl2 + H2 b. CH4 + 2O2 _____
MA_775_DIABLO [31]

Answer:

a.

  • 1 mol de Hierro reacciona con 2 moles de acido clorhidrico para formar 1 mol de cloruro de hierro (II)
  • 6.02×10²³ moleculas de hierro reaccionan con 1.20×10²⁴ moleculas de acido clorhidrico para formar 6.02×10²³ moleculas de cloruro de hierro (II)
  • 55.85 g de hierro reaccionan con 72.9 gramos de acido clorhidrico para formar 126.75 g de cloruro férrico.

b.

  • 6.02×10²³ moleculas de metano reaccionan con 1.20×10²⁴ moleculas de oxigeno para formar 6.02×10²³ moleculas de dioxido de carbono y 1.20×10²⁴ moleculas de agua.
  • 1 mol de metano reacciona con dos moles de oxigeno para generar 1 mol de dióxido de carbono y dos moles de agua en estado de vapor.
  • 16 gramos de metano reaccionan con 64 g de oxigeno para formar 44 gramos de dioxido de carbono y 36 gramos de agua.

c.

  • 3 moles de plata sólida reaccionan con 4 moles de acido nitrico para formar 3 moles de nitrato de plata, 1 mol de monoxido de nitrogeno y 2 moles de agua.
  • 1.80×10²⁴ moleculas de plata reaccionan con 2.41×10²⁴ moleculas de acido nitrico para formar 1.80×10²⁴ moleculas de nitrato de plata, 6.02×10²³ moleculas de monoxido de nitrogeno y 1.20×10²⁴ moleculas de agua.
  • 323.58 g de plata reaccionan con 252 g de acido nitrico para formar 509.58 g de nitrato de plata, 30 g de monoxido de nitrogeno y 36 g de agua.

Explanation:

a. Fe (s)  +2 HCl (aq) → FeCl₂ (aq)

1 mol de Hierro reacciona con 2 moles de acido clorhidrico para formar 1 mol de cloruro de hierro (II)

Calculamos cuanto son dos moles de moleculas sabiendo que:

6.02×10²³ moleculas / 1 mol  . 2 mol = 1.20×10²⁴ moleculas. Entonces

6.02×10²³ moleculas de hierro reaccionan con 1.20×10²⁴ moleculas de acido clorhidrico para formar 6.02×10²³ moleculas de cloruro de hierro (II)

Calculamos las masas molares de cada reactivo y producto

Fe = 55.85 g

HCl = 36.45 g

FeCl₂ = 126.75 g

55.85 g de hierro reaccionan con 72.9 gramos de acido clorhidrico para formar 126.75 g de cloruro férrico.

b. CH₄(g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

1 mol de metano reacciona con dos moles de oxigeno para generar 1 mol de dióxido de carbono y dos moles de agua en estado de vapor.

6.02×10²³ moleculas de metano reaccionan con 1.20×10²⁴ moleculas de oxigeno para formar 6.02×10²³ moleculas de dioxido de carbono y 1.20×10²⁴ moleculas de agua.

Calculamos las masas molares:

CH₄ = 16 g

O₂ = 32 g

CO₂ = 44 g

H₂O g = 18 g

16 gramos de metano reaccionan con 64 g de oxigeno para formar 44 gramos de dioxido de carbono y 36 gramos de agua.

c. 3 Ag (s) + 4HNO3 (aq) → 3 AgNO3 (aq) + NO (g) + 2H₂O (aq)

Calculamos cuantos moleculas contienen 3 y 4 moles:

6.02×10²³  . 3 = 1.80×10²⁴ moleculas

6.02×10²³  . 4 = 2.41×10²⁴ moleculas

3 moles de plata sólida reaccionan con 4 moles de acido nitrico para formar 3 moles de nitrato de plata, 1 mol de monoxido de nitrogeno y 2 moles de agua.

1.80×10²⁴ moleculas de plata reaccionan con 2.41×10²⁴ moleculas de acido nitrico para formar 1.80×10²⁴ moleculas de nitrato de plata, 6.02×10²³ moleculas de monoxido de nitrogeno y 1.20×10²⁴ moleculas de agua.

Calcualmos las masas molares:

Ag = 107.86 g

HNO₃ = 63 g

AgNO₃ = 169.86 g

NO = 30 g

H₂O = 18 g

323.58 g de plata reaccionan con 252 g de acido nitrico para formar 509.58 g de nitrato de plata, 30 g de monoxido de nitrogeno y 36 g de agua.

6 0
3 years ago
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