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3 years ago

1.what is meant by good conductor of heat?give 2 examples of good conductors of heat.

Chemistry

1 answer:

Hitman42 [59]3 years ago

6 0

Answer:

Something that conducts heat very well. Two examples are Aluminum and copper

Explanation:

They heat up very quickly

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Assume that 25 ml of 0.440M NaCl are added to 25 ml of 0.320M AgNO3.

Genrish500 [490]

The balanced chemical equation would be as follows:

NaCl + AgNO3 -> NaNO3 + AgCl

We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:

0.0440 mol/L NaCl (.025 L) = 0.0011 mol NaCl -----> consumed completely and therefore the limiting reactant
0.320 mol/L AgNO3 (0.025 L) = 0.008 mol AgNO3

0.0011 mol NaCl ( 1 mol AgCl / 1 mol NaCl) = 0.0011 AgCl precipitate produced

8 0

3 years ago

The lewis structure of pf3 shows that the central phosphorus atom has ________ nonbonding and ________ bonding electron pair(s)

vfiekz [6]

The Lewis structure of PF3 shows that the central phosphorus atom has one non bonding and three bonding electron pairs. In this compound each Phosphorus atom contributes 5 valence electrons while each fluoride contributes 7 valence electrons making a total of 26 valence electrons. The central Phosphorus atom forms single bonds with each of the fluoride atoms. Phosphorus, therefore ends up with a non-bonding pair since the fluoride atoms already have 8 electrons around them.

6 0

3 years ago

Read 2 more answers

A gas in a sealed container has an initial pressure of 125 kPa at 25.0∘C. If the pressure is increased to 150.0 kPa, what will t

Serjik [45]

Answer:

$T'=30^{\circ} C$

Explanation:

Initial pressure, P = 125 kPa

Initial temperature, T = 25.0°C

If new pressure is increased to P' = 150.0 kPa

We need to find new temperature. According to Gay-Lussac's law,

P ∝ T

$\dfrac{P}{T}=\dfrac{P'}{T'}\\\\T'=\dfrac{P'T}{P}\\\\T'=\dfrac{150\times 25}{125}\\\\=30^{\circ} C$

So, the new temperature will be $30^{\circ} C$ .

5 0

3 years ago

How many electrons does a single oxygen gain or lose in the following reaction?

REY [17]

Answer:

it gains 2 electrons

Explanation:

atoms want to be more stable, for oxygen it's number is 8, 2 less than the stable ring of 10. and hydrogen has a single electron, 1 less than the stable ring of 2. so oxygen takes 2 electrons 2 make it stable and hydrogen becomes attatched to the oxygen atom to form a covalent bond

8 0

3 years ago

Read 2 more answers

Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 2.55 g

Charra [1.4K]

Answer:

4.71 g of sodium and 7.25 g of chlorine

Explanation:

Note: The question is incomplete. the complete question is given below:

Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 2.55 g of sodium and 3.93 g of chlorine. Being consistent with the law of constant composition, also called the law of definite proportions or law of definite composition, which set of masses could be the result of the decomposition of the other sample?

4.71 g of sodium and 3.30 g of chlorine

4.71 g of sodium and 7.25 g of chlorine

4.71 g of sodium and 1.31 g of chlorine

4.71 g of sodium and 13.7 g of chlorine

The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements combined in the same proportion by mass.

This means that irrespective of the source of any sample of a pure chemical compound, the constituents elements are always combined in the same mass ratio.

In the first sample, the mass ratio of Sodium to chlorine is given below:

mass of sodium = 2.55 g

mass of chlorine = 3.93 g

mass ratio; sodium to chlorine = 2.55 / 3.93 = 0.65

From the set of masses give above, we can determine the result of the decomposition of the second sample.

4.71 g of sodium and 3.30 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 3.30 = 1.43

4.71 g of sodium and 7.25 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 7.25 = 0.65

4.71 g of sodium and 1.31 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 1.31 = 3.59

4.71 g of sodium and 13.7 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 13.7 = 0.34

From the results above, the correct set of masses for the second sample is 4.71 g of sodium and 7.25 g of chlorine

3 0

3 years ago