Answer:
The answers to your question are below
Explanation:
a) 6.85×1020 H2O2 molecules
H2O2 MW = 32 + 2 = 34 g
34g -------------------- 6.023 x 10²³ molecules
x ------------------- 6.85 x 10 ²⁰
x = (6.85 x 10 ²⁰)(34)/ 6.023 x 10²³
x = 0.038 g
3.3×1022 SO2 molecules
MW SO2 = 32 + 32 = 64g
64 g -------------------- 6.023 x 10²³ molecules
x -------------------- 3.3×1022 SO2 molecules
x = (3.3×1022 SO2)(64) / 6.023 x 10²³
x = 3.51 g
5.5×1025 O3 molecules
MW = 16 x 3 = 48g
48 g ----------------- 6.023 x 10²³ molecules
x ------------------ 5.5×1025 O3 molecules
x = (5.5×1025 )(48) / 6.023 x 10²³
x = 4383 g
9.30×1019 CH4 molecules
MW = 12 + 4 = 16 g
16 g -------------------- 6.023 x 10²³ molecules
x -------------------- 9.30×1019 CH4 molecules
x = (9.30×1019)(16) / 6.023 x 10²³
x = 0.0025 g
Explanation:
to set 1 container inside, without air movement. 1 outside in that location. compare the 2 containers to see which container has less or more fluid...
Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3
THE MOLECULE HAS A C=C AND AN -OH GROUP, SO IT IS CALLED AN ENE/OL, I.E., AN ENOL. ENOLS CAN BE FORMED ONLY FROM CARBONYL COMPOUNDS WHICH HAVE ALPHA HYDROGENS. THEY CAN BE FORMED BY ACID OR BASE CATALYSIS, AND ONCE FORMED ARE HIGHLY REACTIVE TOWARD ELECTROPHILES, LIKE BROMINE.
