<span>it is located directly under the sima</span>
Mass of PH3= 6.086 g
<h3>Further explanation</h3>
Given
6.0 L of H2
Required
mass of PH3
Solution
Reaction
P4 + 6H2 → 4PH3
Assumed at STP ( 1 mol gas=22.4 L)
Mol of H2 for 6 L :
= 6 : 22.4 L
= 0.268
From the equation, mol PH3 :
= 4/6 x moles H2
= 4/6 x 0.268
= 0.179
Mass PH3 :
= 0.179 x 33,99758 g/mol
= 6.086 g
"RED" color bends the least when passing through a prism.
Answer:
The correct answer is option A.
Explanation:
Volumetric flask : A glass ware with round lower body with flat bottom and with thin cylindrical neck along with mark which indicates the specific volume filled to that mark.It is used in preparation of standard solution of compound with desired concentration with fixed volume.
Erlenmeyer flask
: is a flask with conical shape with flat bottom used in titration experiments to carry out reaction with fixed volume of solution.
Test tube
: Small cylindrical tube with rounded bottom used to observe reaction in between reactant taken in small amount.
Graduated beaker
: Laboratory glassware used measure larger volumes of solution or to mix or stir solutions and liquids.
Graduated cylinder : Laboratory thin cylindrical glassware with accurate marking of volume used to measure an accurate volume of solutions or liquids required in an experiment.
<em><u>Volumetric flask</u></em> is the best piece of laboratory glassware for preparing 500.0 mL of an aqueous solution of a solid
Answer:
a. 1810mL
Explanation:
When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:
where the temperatures must be measured in Kelvin
To convert from Celsius to Kelvin, add 273, or use the equation: 
For this problem, one must also recall that standard temperature is 0°C (or 273K).
So, ![T_1 = 273[K]](https://tex.z-dn.net/?f=T_1%20%3D%20273%5BK%5D)
, and ![T_2 = (49.4+273)[K]=322.4[K]](https://tex.z-dn.net/?f=T_2%20%3D%20%2849.4%2B273%29%5BK%5D%3D322.4%5BK%5D)
.
![\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5D%29%7D%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%29%7D)
![\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%29%7D%28322.4%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%20%29%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29%7D%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29)
![1810.04571428[mL]=V_2](https://tex.z-dn.net/?f=1810.04571428%5BmL%5D%3DV_2)
Adjusting for significant figures, this gives ![V_2=1810[mL]](https://tex.z-dn.net/?f=V_2%3D1810%5BmL%5D)
