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Anni [7]
3 years ago
14

What did Dalton determine about the relationship between atoms, elements, and compounds?

Chemistry
2 answers:
RideAnS [48]3 years ago
7 0

Answer: got on xbox

Explanation:

kirill115 [55]3 years ago
6 0

Answer:

Usman I’m trolling you right know

Explanation:

Get on xbox

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thermometer, calorimeter

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HCI + NaOH --> NaCl + H 20
Vikki [24]

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option D is correct

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6 0
2 years ago
1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation
Rasek [7]

Answer:

Q1: 728.6 J.

Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

<em>Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:</em>

  • The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

<em>Q2:  Calculate the energy change (q) of the surroundings (water) using the enthalpy equation </em>

<em>qwater = m × c × ΔT.  </em>

<em>We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.</em>

<em></em>

a) First part: the energy change (q) of the surroundings (water):

  • The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (25 g)(4.18 J/g°C)(6.4°C) = <em>668.8 J.</em>

<em>b) second part:</em>

<em>Q water = Q unknown metal. </em>

<em>Q unknown metal =  - </em>668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

<em>Q unknown metal =  - </em>668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

<em>- </em>668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (<em>- </em>668.8)/(- 1914) = 0.3495 J/g°C.

<em></em>

3 0
3 years ago
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