It is B. Thank you later please and do good on the test!
Answer:
120g
Explanation:
Step 1:
We'll begin by writing the balanced equation for the reaction.
Sn + 2HF —> SnF2 + H2
Step 2:
Determination of the number of mole HF needed to react with 3 moles of Sn.
From the balanced equation above,
1 mole of Sn and reacted with 2 moles of HF.
Therefore, 3 moles Sn will react with = 3 x 2 = 6 moles of HF.
Step 3:
Conversion of 6 moles of HF to grams.
Number of mole HF = 6 moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn.
Answer:
1. 0.178 moles ; 2. 8x10²³ atoms ; 3. 7.22x10²³ molecules ; 4. 89.6 g ; 5. 1.34x10²² atoms ; 6. 1.67x10²⁵ molecules
Explanation:
1. Mass / Molar mass = Mol
5g / 28 g/m = 0.178 moles
2. 1 molecule of N₂ has 2 atoms, it is a dyatomic molecule.
4x10²³ x2 = 8x10²³ atoms
3. 1 mol of anything, has 6.02x10²³ particles
6.02x10²³ molecules . 1.2 mol = 7.22x10²³
4. 1 atom of C weighs 12 amu.
4.5x10²⁴ weigh ( 4.5x10²⁴ . 12) = 5.24x10²⁵ amu
1 amu = 1.66054x10⁻²⁴g
5.24x10²⁵ amu = (5.24x10²⁵ . 1.66054x10⁻²⁴) = 89.6 g
5. Molar mass NaCl = 58.45 g/m
1.3 g / 58.45 g/m = 0.0222 moles
1 mol has 6.02x10²³ atoms
0.0222 moles → ( 0.0222 . 6.02x10²³) = 1.34x10²²
6. Density of water is 1 g/mL, so 500 mL are contained in 500 g of water
Molar mass H₂O = 18 g/m
500 g / 18 g/m = 27.8 moles
6.02x10²³ molecules . 27.8 moles = 1.67x10²⁵
Mineral composition affects the classification of igneous rock. in simplified classification, igneous to is are classified by the type of feldspar present, by the I type a of feldspar present, or the absence of quartz. in case of neither present, then by the type of iron or magnesium present.
Answer:
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
Explanation:
The pH of the solution = 13.00
pH + pOH = 14
pOH = 14 - pH = 14 - 13.00 = 1.00
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![1.00=-\log[OH^-]](https://tex.z-dn.net/?f=1.00%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-1.00} M=0.100 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-1.00%7D%20M%3D0.100%20M)
![[KOH]=[OH^-]=[K^+]=0.100 M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5BK%5E%2B%5D%3D0.100%20M)
Molariy of the KOH = 0.100 M
Volume of the KOH solution = 800 mL= 0.800 L
1 mL = 0.001 L
Moles of KOH = n
n = 0.0800 mol
Mass of 0.0800 moles of KOH :
0.0800 mol × 56 g/mol = 4.48 g
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
