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Romashka [77]
2 years ago
6

A 90.0 g sample of an unknown metal absorbed 25.6 J of heat as its temperature increased 1.18 degrees C. What is the specific he

at of the metal
Chemistry
1 answer:
jenyasd209 [6]2 years ago
8 0

The specific heat of the metal is 241.05 J/kg°C

<h3>What is specific heat capacity?</h3>

Specific heat capacity can be defined as the amount of heat required to raise a unit mass of metal through 1 °C

To calculate the specific heat capacity of the metal, we use the formula below.

Formula:

  • Q = cm(t'-t)............... Equation 1

Where:

  • Q = Heat absorbed by the metal
  • c = specific heat capacity of the metal
  • m = mass of the metal
  • Δt = increase in temperature

make c the subject of the equation

  • c = Q/m(Δt)................. Equation 2

From the question,

Given:

  • Q = 25.6 J
  • m = 90 g = 0.09 kg
  • Δt = 1.18 °C

Substitute these values into equation 2

  • c = 25.6/(1.18×0.09)
  • c = 241.05 J/kg°C

Hence, the specific heat capacity of the metal is 241.05 J/kg°C.

Learn more about specific heat capacity here: brainly.com/question/21406849

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Deuterium is a relatively uncommon form of hydrogen, but can be created from what common source?.
Elena-2011 [213]

Deuterium is a relatively uncommon form of hydrogen, but can be created from water.

  • Heavy hydrogen commonly  known as deuterium
  • stable isotopes of hydrogen
  • gets its name from the Greek word deuterons means second.
  • has only one proton and one neutron
  • nucleus of the hydrogen's deuterium atom is known as a deuteron containing one proton and one neutron.
  • Deuterium forms chemical bonds that are stronger than regular hydrogen
  • gas deuterium is colorless
  • Deuterated water is used in Magnetic Resonance Spectroscopy.
  • used in the determination of the isotopologue of various organic compounds.
  • used in Infrared Spectroscopy.

To know more about Deuterium visit : brainly.com/question/27870183

#SPJ4

8 0
1 year ago
If a student mixes 75 mL of 1.30 M HNO3 and 150 mL of 6.5 M NaOH. is the final solution acidic, basic, or neutral
raketka [301]

Answer:

The solution is basic.

Explanation:

We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):

  • If no. of millimoles of acid > that of base; the solution is acidic.
  • If no. of millimoles of acid = that of base; the solution is neutral.
  • If no. of millimoles of acid < that of base; the solution is basic.

  • We need to calculate the no. of millimoles of acid and base:

no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.

no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.

<em>∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).</em>

<em>So, the solution is: basic.</em>

4 0
2 years ago
1.
dolphi86 [110]

Answer: D. transverse

Explanation:

Light is a transverse wave, while sound is a longitudinal wave.

8 0
2 years ago
Three compounds that contain elements from Group 16 are SeO2F2 , SeOF2,
egoroff_w [7]

Answer:

A

Explanation:

A formal charge (FC) is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity(Wikipedia).

The formal charge on an atom in a molecule reflects the electron count associated with the atom compared to the isolated neutral atom(University of Calgary).

Looking at all the structures listed A-E for SeO2F2, only structure A minimizes the formal charges for each atom in SeO2F2.

4 0
2 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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