Answer:
mass of hydrogen collected is 0.016 gram
Explanation:
Given values:
For calculating mass we have to find the number of moles first
Ideal gas equation PV =nRT
Volume= 195 ml
Pressure: 753 torr =0.99 atm
Temperature: 25+273= 298 Kelvin
Ideal gas constant R= 0.0821 Latm/molK
Number of moles n= ?
So n= PV/RT
Adding the values
n= = (753 torr)(1 atm/760 torr)(195 mL)(1 L/1000 mL)/(0.0821 L·atm/mol·K)(273 K)
n = 0.007897 moles of H₂
Now mass of hydrogen collected = number of moles x Molar mass of H₂
= 0.007897 x 2
= 0.0157 g H₂
Now mass of hydrogen collected is 0.016 gram (rounding the amount)
Answer:
2.9g/cm^3
Explanation:
We're given that the mass of the system before the rock was added was
120g
. The mass after the rock was added was 250g, so the rock's mass is 130g.
The initial volume of water in the graduated cylinder was
30
mL
, and the volume after the rock was added read 75mL
, so the rock's volume is 45mL
Density = Mass over Volume.
AKA. 130g/45mL
2.9g/cm^3 (rounded)
I would say A. Because it looks as though it’s the best answer
Answer:
<u>Molar</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>diprotic</u><u> </u><u>acid</u><u> </u><u>is</u><u> </u><u>4</u><u>2</u><u>4</u><u> </u><u>grams</u>
Explanation:
[hint: <u>diprotic</u><u> </u><u>acid</u><u> </u><u>only</u><u> </u><u>contains</u><u> </u><u>2</u><u> </u><u>hydrogen</u><u> </u><u>protons</u><u>]</u>
Ionic equation:

first, we get moles of potassium hydroxide in 28.94 ml :

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.
Therefore, <u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>c</u><u>i</u><u>d</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>r</u><u>e</u><u>a</u><u>c</u><u>t</u><u>e</u><u>d</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>8</u><u>7</u><u>4</u><u>3</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u>.</u>

for the molar mass: