<span> Pb(NO3)2 + 2 NaBr ------> PbBr2 + 2 NaNO3
2 mol 2 mol
M(NaBr)= 102 g/mol
M(NaNO3) = 85 g/mol
1) 244g NaNO3 *1 mol NaNO3/85 g NaNO3 = 244/85 mol NaNO3
2)</span> Pb(NO3)2 + 2 NaBr ------> PbBr2 + 2 NaNO3
2 mol 2 mol<span>
x mol </span>244/85 mol
<span>
x=(2 mol*</span> 244/85 mol )/2 mol = 244/85 mol NaBr
<span>
3) </span> 244/85 mol NaBr*102g NaBr/1 mol = (244*102/85) g NaBr =292.8 g NaBr<span>
</span>
Answer:
Cr₂S₃
Explanation:
From the question given above, the following data were obtained:
Mass of chromium (Cr) = 0.67 g
Mass of chromium sulfide = 1.2888 g
Empirical formula =?
Next, we shall determine the mass of sulphur (S) in the compound. This can be obtained as follow:
Mass of chromium (Cr) = 0.67 g
Mass of chromium sulfide = 1.2888 g
Mass of sulphur (S) =?
Mass of S = (Mass of chromium sulfide) – (Mass of Cr)
Mass of S = 1.2888 – 0.67
Mass of S = 0.6188 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Cr = 0.67 g
Mass of S = 0.6188 g
Divide by their molar mass
Cr = 0.67 / 52 = 0.013
S = 0.6188 / 32 = 0.019
Divide by the smallest
Cr = 0.013 / 0.013 = 1
S = 0.019 / 0.013 = 1.46
Multiply by 2 to express in whole number
Cr = 1 × 2 = 2
S = 1.46 × 2 = 3
Therefore, the empirical formula of the compound is Cr₂S₃
<span>The electron transport process makes water and ATP and is sometimes called Oxidative phosphorylation because it requires oxygen.</span>
Mm of Hg is the same as Torres, therefore, it should be 105.6 Torr
Answer:
2.48 g
Explanation:
From the question given above, the following data were obtained:
Number of atoms = 2.53×10³⁴ atoms
Mass of Ni =?
From Avogadro's hypothesis,
we understood that 1 mole of any substance contains 6.02×10²³ atoms.
Therefore, 1 mole of Ni contains 6.02×10²³ atoms.
Recall:
1 mole Ni = 59 g
Therefore, 59 g of Ni contains 6.02×10²³ atoms.
Finally, we shall determine the mass of 2.53×10³⁴ atoms of Ni. This can be obtained as follow:
6.02×10²³ atoms = 59 g
Therefore,
2.53×10³⁴ = 2.53×10³⁴ × 59 / 6.02×10²³
2.53×10³⁴ atoms = 2.48 g
Thus, the mass of 2.53×10³⁴ atoms of Ni is 2.48 g.
