when 4.5 mol Al react with 11.5 mol HCI, what is the limiting reactant and how many moles of H2 can be formed?

The men's world record for swimming 1500.0 m in a long course pool (as of 2007) is 14 min 34.56 s. At this rate, how many second

vesna_86 [32]

Answer:

It will take 140.76 seconds to swim 0.150 miles.

Explanation:

Distance covered by man by swimming for t time= d

d = 1500.0 miles

t = 14 minutes + 34.56 seconds

1 min = 60 seconds

t = 14 × 60 seconds + 34.56 seconds = 874.56 seconds

Speed of the man while swimming = S = $\frac{1500.0 m}{874.56 s}=1.715 m/s$

If the man swims 0.150 miles with the same speed ,then the time taken by him be T.

0.150 miles = 0.150 × 1609.34 m = 241.401 m

(1 mile= 1609.34 meters)

$S=\frac{Distance}{time}=\frac{241.401 m}{T}$

$1.715 m/s=\frac{241.401 m}{T}$

T = 140.76 seconds

It will take 140.76 seconds to swim 0.150 miles.

6 0

3 years ago

. I am described as being conductive, malleable, shiny, and reactive with HCl. I am one of the most reactive elements. I turn bl

mylen [45]

Answer:

Potassium

Explanation:

Most reactive metal in the reactivity series

8 0

3 years ago

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). suppose 8.30 ml of toluene (d = 0.867

GrogVix [38]

Missing question:<span> How many grams of oxygen are needed for complete combustion of the toluene?
Chemical reaction: </span>C₇H₈ + 9O₂ → 7CO₂ + 4H₂<span>O.
V(</span>C₇H₈) = 8,30 ml.
d(C₇H₈) = 0,867 g/ml.
n(C₇H₈) = V(C₇H₈) · d(C₇H₈).
m(C₇H₈) = 8,3 ml · 0,867 g/ml.
m(C₇H₈) = 7,19 g.
n(C₇H₈) = m(C₇H₈) ÷ M(C₇H₈).
n(C₇H₈) = 7,19 g ÷ 92,14 g/mol = 0,078 mol.
From chemical reaction: n(C₇H₈) : n(O₂) = 1 : 9.
n(O₂) = 9 · 0,078 mol = 0,702 mol.
m(O₂) = 0,702 mol · 32 g/mol = 22,5 g.

3 0

3 years ago

What is ppm?

elena-s [515]

Answer:

What does ppm mean? This is an abbreviation for "parts per million" and it also can be expressed as milligrams per liter (mg/L). This measurement is the mass of a chemical or contaminate per unit volume of water. Seeing ppm or mg/L on a lab report means the same thing.

that is all i have to say.

4 0

3 years ago

How many liters of hydrogen gas is produced from 3.712 g of magnesium with 104.2ml of 1.385 mol/L HCL (aq) at SATP? Please show

Licemer1 [7]

Answer:

$V=1.61L$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$Mg+2HCl\rightarrow MgCl_2+H_2$

Next, we compute the reacting moles of each reactants:

$n_{Mg}=3.712gMg*\frac{1molMg}{24.305 gMg}=0.153molMg$

$n_{HCl}=1.385\frac{molHCl}{L}*0.1042L=0.144molHCl$

Then, as magnesium and hydrohloric acid are in a 1:2 molar ratio 0.153 moles of magnesium will completely react with 0.306 moles of hydrochloric acid yet we only have 0.144 moles, therefore, limiting reactant is hydrochloric acid. Thus, we compute the produced moles of hydrogen:

$n_{H_2}=0.144molHCl*\frac{1molH_2}{2molHCl} =0.072molH_2$

Finally, we use the ideal gas equation with T=298K and 1atm (STP conditions) to compute the liters of hydrogen gas:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{0.072mol*0.082\frac{atm*L}{mol*K}*273K}{1atm}\\ \\V=1.61L$

Best regards.

3 0

3 years ago