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ivanzaharov [21]
3 years ago
13

Which scientist is credited with developing the fist scientific atomic theory

Chemistry
1 answer:
Norma-Jean [14]3 years ago
3 0

Answer:

German chemist G.E. Stahl

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Find the number of CoCl2 units present in a 0.78 mol sample.
stich3 [128]

\qquad\qquad\huge\underline{{\sf Answer}}

Here we go ~

1 mole of \sf{ COCl_2} has 6.022 × 10²³ molecules of the given compound.

So, 0.78 mole of \sf{COCl_2} will have ~

\qquad \sf  \dashrightarrow \: 0.78 \times 6.022 \times 10 {}^{23}

\qquad \sf  \dashrightarrow \:  \approx4.7 \times 10 {}^{23}  \:  \:  \: molecules

5 0
2 years ago
Trình bày sự hình thành phân tử CCl4 (lai hoá sp3)
Anna35 [415]
Gucgjfgdjgcycc fh(xsjfhghjkm
8 0
3 years ago
Determine the number of moles of a aluminum sample containing 2.89E26 atoms.
Brut [27]

Answer:

Explanation:

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8 0
3 years ago
Heating curve shows temperature verses energy gain. Which parts of the curve represent a gain in potential energy?
Brilliant_brown [7]

Answer:

Those two horizontal lines.

Explanation:

Hello there!

In this case, when focusing on these heating curves, it is important to say they tend to have two constant-temperature sections and three variable-temperature sections. Thus, from lower to higher temperature, the first constant-temperature section corresponds to melting and the second one vaporization, whereas the three variable-temperature sections correspond to the heating of the solid until melting, the liquid until vaporization and the gas until the critical point.

In such a way, we infer that the boxes referred to constant temperature are referred to a gain in potential energy, that is, the two horizontal lines.

Regards!

6 0
3 years ago
A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe
kaheart [24]

Answer:

The specific heat of the alloy C_{a} = 0.37 \frac{KJ}{Kg K}

Explanation:

Mass of an alloy m_{a} = 25 gm

Initial temperature T_{a} = 100°c = 373 K

Mass of water m_{w} = 90 gm

Initial temperature of water T_{w} = 25.32 °c = 298.32 K

Final temperature T_{f} = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

m_{a} C_{a}  [T_{a} - T_{f}] = m_{w} C_w (T_{f} -T_{w} )

25 × C_{a} × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

C_{a} = 0.37 \frac{KJ}{Kg K}

This is the specific heat of the alloy.

6 0
3 years ago
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