Molar mass :
NaBr = 103 g/mol
Pb(NO3)2 = 331.20 g/mol
<span><span /><span>Balanced chemical equation :
</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
</span><span>
2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
g NaBr -------------------> 311 g Pb(NO3)2
331.20 g = 2*103*311
331.20 g = 64066
mass ( NaBr ) = 64066 / 331.20
mass ( naBr) = 193,43 g of NaBr
hope this helps!.
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Answer:
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Explanation:
1) Balanced chemical equation:
2SO2 (g) + O2 (g) -> 2SO3 (l)
2) Molar ratios
2 mol SO2 : 1 mol O2 : 2 mol SO3
3) Convert 6.00 g O2 to moles
number of moles = mass in grams / molar mass
number of moles = 6.00 g / 32 g/mol = 0.1875 mol O2.
4) Use proportions with the molar ratios
=> 2 moles SO2 / 1 mol O2 = x / 0.1875 mol O2
=> x = 0.1875 mol O2 * 2 mol SO2 / 1 mol O2 = 0.375 mol SO2.
5) Convert 0.375 mol SO2 to grams
mass in grams = number of moles * molar mass
molar mass SO2 = 32 g/mol + 2*16 g/mol = 64 g/mol
=> mass SO2 = 0.375 mol * 64 g / mol = 24.0 g
Answer: 24.0 g of SO2 are needed to react completely with 6.00 g O2.
Answer:
0.1
Explanation:
We must first put down the equation of the reaction in order to guide our solution of the question.
2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)
Now from the question, the following were given;
Concentration of acid CA= ??????
Concentration of base CB= 0.299M
Volume of acid VA= 17.8ml
Volume of base VB= 24.7ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB= NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
SUBSTITUTING VALUES;
CA= 0.299 × 24.7 ×2 / 17.8×1
CA= 0.8298 M
But;
pH= -log[H^+]
[H^+] = 0.8298 M
pH= -log[0.8298 M]
pH= 0.1
