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MaRussiya [10]
2 years ago
6

Is gas occupies 733 cm at 10.9 at what temp will it occupy 950 cm

Chemistry
1 answer:
alexgriva [62]2 years ago
5 0

Temperature of gas at volume of 950 cm³ is 368.14K or 94.99⁰c.

  • Volume is the amount of space a three-dimensional object takes up, expressed in cubic units.
  • units of volume are mL, liter, cm³ or m³.

Given,

in this question, volume occupied by gas at 10.9⁰c is 733cm³.

we have to find out temperature at which gas occupies volume of 950 cm³.

First, convert temperature from celcius to Kelvin

10.9⁰c = 10.9 + 273.15 = 284.05 K

By equation or Boyle's Law

P1 x V1 / T1 = P2 x V2 / T2

Pressure is constant here

therefore, V1 / T1 = V2 / T2

here, V1 = 733 cm³

T1 = 284.05 K

V2 = 950 cm³

we have to find out T2

So, 733 / 284.05 = 950 / T2

T2 = 284.05 × 950 / 733 = 368.14 K

= 94. 99⁰c

Therefore the temperature of gas at volume of 950 cm³ is 368.14K or 94.99⁰c.

Learn more about Boyle's Law here:

https://brainly.in/question/2900894

#SPJ9

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10 points please please help
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4 0
3 years ago
What is the mass of 2.90 ×1022 molecules of NaOH (Molar mass = 40.0 g/mol)?
ivann1987 [24]
40.0 g ( 1 mole ) --------------- 6.02x10²³ molecules
      ? ? --------------------------- 2.90x10²² molecules

mass = 2.90x10²² * 40.0 /  6.02x10²³

mass = 1.16x10²⁴ / 6.02x10²³

mass = 1.9269 g

hope this helps!

3 0
3 years ago
Hiii pls help me to write out the ionic equation ​
emmasim [6.3K]

Answer:

<u>STEP I</u>

This is the balanced equation for the given reaction:-

2KOH_{(aq)} + H_2SO_4{}_{(aq)}   \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}

<u>STEP II</u>

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

  • 2KOH -> 2K+ + 2OH-
  • H2SO4 -> 2H+ + (SO4)2-
  • K2SO4 -> 2K+ + (SO4)2-

<u>STEP III</u>

Rewriting these in the form of equation

\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \:  \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O

<u>STEP </u><u>IV</u>

Canceling spectator ions, the ions that appear the same on either side of the equation

<em>(note: in the above step the ions in bold have gotten canceled.)</em>

\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}

This is the net ionic equation.

____________________________

\\

\mathfrak{\underline{\green{ Why\: KOH \:has\:  been\: taken\: as\: aqueous ?}}}

  • KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

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Answer:

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Explanation:

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