I think sulfuric acid cannot be regarded as an ampholyte. Ampholytes are amphoteric molecules or that contains both acidic and basic groups. An amphoteric compound can react both as an acid and a base which sulfuric acid is not capable.
Answer:
There is a temperature of 111 K required.
Explanation:
Step 1: Data given
Volume of hydrogen gas = 6.34 L
Pressure = 1.52 atm at 298 K
The volume of the gas at a pressure of 0.720 atm is 4.98 L
Step 2:
(P1*V1)/T1 = (P2*V2)/T2
⇒ with P1 = the pressure of the hydrogen gas = 1.52 atm
⇒ with V1 = the volume of the hydrogen gas = 6.34 L
⇒ with T1 = the temperature of the hydrogen gas = 298 K
⇒ with P2 = the changed pressure of the gas = 0.720 atm
⇒ with V2 = the changed volume of the gas = 4.98 L
⇒ with T2= the temperature of the gas at 0.720 atm and 4.98L
(1.52*6.34)/298 = (0.720*4.98)/T2
0.0323 = 3.5856/T2
T2 = 110.88 K ≈111K
There is a temperature of 111 K required.
Answer:
![\boxed {\boxed {\sf A. \ 2.48 \ mol \ Na_2O}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%20A.%20%5C%202.48%20%5C%20mol%20%5C%20Na_2O%7D%7D)
Explanation:
We are asked to convert an amount in grams to moles. To do this, we use the molar mass. This is the number of grams in one mole of a substance. It is the same value numerically as the atomic mass on the Periodic Table, however the units are grams per mole, not atomic mass units.
Look up the molar masses for the individual elements.
- Sodium (Na): 22.9897693 g/mol
- Oxygen (O): 15.999 g/mol
Look back at the formula: Na₂O. Notice there is a subscript of 2 after sodium. This means there are 2 atoms of sodium in every molecule, so we have to multiply sodium's molar mass by 2 before adding oxygen's.
- Na₂O: 2(22.9897693 g/mol)+ 15.999 g/mol = 61.9785386 g/mol
Set up a ratio using the molar mass.
![\frac {61.9785386 \ g \ Na_2O}{1 \ mol \ Na_2O}](https://tex.z-dn.net/?f=%5Cfrac%20%7B61.9785386%20%5C%20g%20%5C%20Na_2O%7D%7B1%20%5C%20mol%20%5C%20Na_2O%7D)
Multiply by the given number of grams.
![154 \ g \ Na_2O*\frac {61.9785386 \ g \ Na_2O}{1 \ mol \ Na_2O}](https://tex.z-dn.net/?f=154%20%5C%20g%20%5C%20Na_2O%2A%5Cfrac%20%7B61.9785386%20%5C%20g%20%5C%20Na_2O%7D%7B1%20%5C%20mol%20%5C%20Na_2O%7D)
Flip the ratio so the grams of sodium oxide can cancel each other out.
![154 \ g \ Na_2O*\frac {1 \ mol \ Na_2O}{61.9785386 \ g \ Na_2O}](https://tex.z-dn.net/?f=154%20%5C%20g%20%5C%20Na_2O%2A%5Cfrac%20%7B1%20%20%5C%20mol%20%5C%20Na_2O%7D%7B61.9785386%20%5C%20g%20%5C%20Na_2O%7D)
![154 *\frac {1 \ mol \ Na_2O}{61.9785386 }](https://tex.z-dn.net/?f=154%20%2A%5Cfrac%20%7B1%20%20%5C%20mol%20%5C%20Na_2O%7D%7B61.9785386%20%7D)
![\frac {154}{61.9785386 } \ mol \ Na_2O](https://tex.z-dn.net/?f=%5Cfrac%20%7B154%7D%7B61.9785386%20%7D%20%20%5C%20mol%20%5C%20Na_2O)
![2.48473106141 \ mol \ Na_2O](https://tex.z-dn.net/?f=2.48473106141%20%5C%20mol%20%5C%20Na_2O)
The original measurement of grams given has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place.
The 4 in the thousandth place tells us to leave the 8.
![2.48 \ mol \ Na_2O](https://tex.z-dn.net/?f=2.48%20%5C%20mol%20%5C%20Na_2O)
There are <u>2.48 moles of sodium oxide</u> in 154 grams, so choice A is correct.
Answer: 110.2 lbs
Explanation: First, let's determine how many kg the patient masses. Since the dose is 3.000 mg/kg we can convert that to 0.003000 g/kg. So
0.1500 g / 0.003000 g/kg = 50 kg.
So we know the patient has a weight of 50 kg. Now we simply multiply by 2.20462 to get lbs. So
50 kg * 2.20462 lbs/kg = 110.231 lbs
Rounding to 1 decimal place gives 110.2 lbs
Answer:
The number on the top left of the element.