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vodomira [7]
3 years ago
9

What do we assume about the volume of the actual molecules themselves in a sample of gas, compared to the bulk volume of the gas

overall? Why?
Chemistry
1 answer:
Firlakuza [10]3 years ago
4 0

We assume that the volume of the molecules themselves in a gas sample is negligible compared to the bulk volume of the gas sample: this helps us to explain why gases are so compressible.

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How to find the number of neutrons in an isotope?
Alexeev081 [22]
Chemical properties of elements are primarily determined by the electrons but not by neutrons. Therefore, the isotopes of the same element have similar chemical behavior. (a) The atomic number is 17, so there are 17 protons and 17 electrons. The mass number is 35, so there are 18 (=35-17) neutrons. Hope this helps. :)
4 0
3 years ago
A hot lump of 30.5 g of iron at an initial temperature of 52.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to r
slava [35]

Answer:

26.7°C

Explanation:

Using the formula; Q = m × c × ΔT

Where; Q = amount of heat

m = mass

c = specific heat

ΔT = change in temperature

In this question involving iron placed into water, the Qwater = Qiron

For water; m= 50g, c = 4.18 J/g°C, Initial temp= 25°C, final temp=?

For iron; m = 30.5g, c = 0.449J/g°C, Initial temp= 52.7°C, final temp=?

Qwater = -(Qiron)

m × c × ΔT (water) =- {m × c × ΔT (iron)}

50 × 4.18 × (T - 25) = - {30.5 × 0.449 × (T - 52.7)}

209 (T - 25) = - {13.6945 (T - 52.7)}

209T - 5225 = -13.6945T + 721.7

209T + 13.6945T = 5225 + 721.7

222.6945T = 5946.7

T = 5946.7/222.6945

T = 26.7

Hence, the final temperature of water and iron is 26.7°C

8 0
3 years ago
How is energy in a washing machine being utilised and wasted?
7nadin3 [17]

Answer:

It depends on whether the setting is hot or cold .-.

Explanation:

8 0
2 years ago
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What is the family and location of Lithium?
motikmotik

Explanation:

\huge{\underline{\underline{\underline{\sf{\pink{ムɳรᏇɛƦ \: ࿐}}}}}}

<h2><em><u>Alkali </u></em><em><u>metal</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em></h2>

<h3><em><u>please</u></em><em><u> mark</u></em><em><u> as</u></em><em><u> brainliest</u></em><em><u> answer</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em></h3>

('\. ./')

( • . • )

(,,) (,,)

5 0
3 years ago
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A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
mestny [16]

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

4 0
3 years ago
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