They are performed in the JEJUNUM
Answer:
The molar concentration of a solution made with 3.744 g of Mg(NO₃)₂ dissolved in enough water to make 50.0 mL of solution is 
Explanation:
Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume.
The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:
In this case:
- Mg: 24.3 g/mole
- N: 14 g/mole
- O: 16 g/mole
So, the molar mass of Mg(NO₃)₂ is:
Mg(NO₃)₂= 24.3 g/mole + 2*(14 g/mole + 3*16 g/mole)= 148.3 g/mole
So, if you have 3.744 g of Mg(NO₃)₂, you can apply the following rule of three: if 148.3 grams of Mg(NO₃)₂ are present in 1 mole, 3.744 grams in how many moles are present?
moles= 0.025
Then you have:
- number of moles=0.025
- volume= 50 mL= 0.05 L (being 1,000 mL= 1 L)
Replacing in the definition of molarity:
you get:
<u><em>The molar concentration of a solution made with 3.744 g of Mg(NO₃)₂ dissolved in enough water to make 50.0 mL of solution is </em></u><u><em></em></u>
If 4 moles of P is used by 5 mole of O2
then....0.489 moles will be used by 5/4 × .489 = .611 moles of O2
so .611 moles
so if 4 moles of P is burnt , 1 mole of P4O10 is produced ....so for .489 moles...... .489/4=.122 moles !
so mass will be .122× 283.89 = 34.7 grams
so first ans is .611 moles and second is 34.7 grams !
if you have any problem regarding this , just comment !!!
Answer: i hope it helps or not
A car increases it's velocity from 0 m/s to 14 m/s in 2 seconds. Amis - Omls ... A racing car's velocity is increased from 44 m/s to 66m/s in 11 seconds. ... 25 km/hr. 1200 km/hr. 2 min. - Kinlarise. 6. A car accelerates from a standstill to 60 km/hr in.
Explanation:
Answer:
The sample will be heated to 808.5 Kelvin
Explanation:
Step 1: Data given
Volume before heating = 2.00L
Temperature before heating = 35.0°C = 308 K
Volume after heating = 5.25 L
Pressure is constant
Step 2: Calculate temperature
V1 / T1 = V2 /T2
⇒ V1 = the initial volume = 2.00 L
⇒ T1 = the initial temperature = 308 K
⇒ V2 = the final volume = 5.25 L
⇒ T2 = The final temperature = TO BE DETERMINED
2.00L / 308.0 = 5.25L / T2
T2 = 5.25/(2.00/308.0)
T2 = 808.5 K
The sample will be heated to 808.5 Kelvin
