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MakcuM [25]
3 years ago
11

Which of the following statements is true about gas particles?

Chemistry
2 answers:
Lynna [10]3 years ago
4 0
<h2>Answer:</h2>

There is <u>a lot of empty space between the gas particles</u>.

<h2>Explanation:</h2>

There are three states of matter called solid, liquid, and gas. Among them gas have high expanding capability. The gaseous particles have large space between them and they lack definite shape or volume. They doesn't repel with other gaseous particle due to strong force.  

As there is huge space between the particles present in the gas, they occupy the whole space of the substance in which they are kept. The strong attracting force between the particles is seen in case of solids due to which they have definite shape and volume.

satela [25.4K]3 years ago
3 0

Answer:

There is a lot of empty space between them

Explanation:

We know that gas molecules are loosely packed,

Therefore there is a lot of intermolecular space...(Which basically means that there is a lot of space between the molecules or particles in a gas)

Happy to help.

Pls mark as Brainliest.

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Write a story of your life when you were hurted by someone whom you trusted blindly...​
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4 0
3 years ago
How many grams of H2O are produced when 35.0 g of NaOH reacts with 17.5 g of CO,?
zhenek [66]

Answer:

2NaOH + CO2 -> Na2CO3 + H2O

1) Find the moles of each substance

\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\  =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles\\

2) Determine the limitting reagent

\\NaOH = \frac{0.8750437522}{2} = 0.4375218761\\\\

∴ Carbon dioxide is limitting as it has a smaller value.

3) multiply the limiting reagent by the mole ratio of unknown over known

n(H2O ) = 0.3976369007 × 1/2

             = 0.1988184504 moles

4) Multiply the number of moles by the molar mass of the substance.

m = 0.1988184504 × (1.008 × 2 + 16.00)

   = 0.1988184504 × 18.016

   = 3.581913202 g

Explanation:

6 0
3 years ago
CaC2(s) + 2H2O(l) --&gt; Ca(OH)2(aq) + C2H2(g) In the reaction above, 0.5487 grams of calcium carbide are completely consumed to
jasenka [17]

Answer:

239.7mL

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas constant (0.0821 Latm/molK)

T = temperature (K)

The balanced chemical equation in this question is as follows:

CaC2(s) + 2H2O(l) --> Ca(OH)2(aq) + C2H2(g)

From the equation, 1 mole of CaC2 produces 1 mole of ethylene gas, C2H2.

Using mole = mass/molar mass

Molar mass of CaC2 = 40 + 12(2)

= 40 + 24

= 64g/mol

mole = 0.5487/64

mole = 0.00857mol of CaC2

Hence, 0.00857mol of CaC2 produced 0.00857mol of C2H2

Based on the information provided, n = 0.00857mol, T = 43°C = 43 + 273 = 316K, p = 0.926 atm

PV = nRT

V = nRT/P

V = 0.00857 × 0.0821 × 316/0.926

V = 0.222/0.926

V = 0.2397L

In mL, volume = 0.2397 × 1000

= 239.7mL

3 0
3 years ago
What does it mean for an acid and base to neutralize each other?
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This means that they mixed .
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