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Katena32 [7]
3 years ago
14

While driving your rental car on your vacation in Europe, you find that you are getting 8.8 km/L of gasoline. What does this val

ue correspond to in miles per gallon?
Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
8 0

Answer : The value correspond to in miles per gallon is, 20.6976 mile/gallon

Explanation :

The conversion used to convert kilometer to miles is:

1 km = 0.6214 miles

The conversion used to convert liter to gallon is:

1 L = 0.2642 gallons

Thus,

1 km/L = \frac{0.6214}{0.2642}mile/gallon

1 km/L = 2.352 mile/gallon

As we are given that 8.8 km/L of gasoline. Now we have to convert it into mile/gallon.

As, 1 km/L = 2.352 mile/gallon

So, 8.8 km/L = \frac{8.8km/L}{1km/L}\times 2.352mile/gallon

                     = 20.6976 mile/gallon

Thus, the value correspond to in miles per gallon is, 20.6976 mile/gallon

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You have 350 mL of 3.4 M hydrochloric acid (HCl). How many grams of HCl gas are dissolved? Bonus: what is the volume of the HCl
Crank

Answer:

1. 43.44g of HCl

2. 26.67 L of HCl

Explanation:

1) Molarity of a solution = number of moles (n) ÷ Volume (V)

According to the provided information in this question,

V = 350 mL = 350/1000 = 0.350L

Molarity = 3.4 M

Using Molarity = n/V

3.4 = n/0.350

n = 3.4 × 0.350

n = 1.19mol

Using the formula below to calculate the mass of HCl;

mole = mass/molar mass

Molar mass of HCl = 1 + 35.5 = 36.5g/mol

mole = mass/MM

mass = 1.19 mol × 36.5g/mol

mass = 43.44g of HCl

2) At STP, HCl has a pressure of 1atm, a temperature of 273K

V = ?

n = 1.19 mol

R = 0.0821 Latm/molK

Using PV = nRT

V = nRT/P

V = 1.19 × 0.0821 × 273/1

Volume = 26.67L

5 0
3 years ago
HELPP! pls
pshichka [43]

Answer:

"2.48 mole" of H₂ are formed. A further explanation is provided below.

Explanation:

The given values are:

Mole of Al,

= 3.22 mole

Mole of HBr,

= 4.96 mole

Now,

(a)

The number of mole of H₂ are:

⇒  \frac{Mole \ of \ H_2}{3} =\frac{Mole \ of HBr}{6}

or,

⇒  Mole \ of \ H_2=\frac{1}{2}\times Mole \ of \ HBr

⇒                      =\frac{1}{2}\times 4.96

⇒                      =2.48 \ mole

(b)

The limiting reactant is:

= HBr

(c)

The excess reactant is:

= Al

6 0
3 years ago
When solid magnesium burns, it reacts with oxygen in the air to form a powder called magnesium oxide. A chemist performed this r
siniylev [52]

Answer:

A. The increase in the mass of the magnesium oxide was due to oxygen atoms in the air.

Explanation:

Burning occurs in the presence of oxygen. A chemical combination occurs between Mg and O in that the atom of magnesium attracts one another.

          Mg + O₂ → MgO

Starting with the magnesium, on reacting with oxygen a new compound forms. This is why there is mass increase in the MgO compared to the starting material.

4 0
3 years ago
Read 2 more answers
Arrange in the correct order of occurrence. Write 1 – 5 for each statement.
ss7ja [257]

Answer:

Explanation:Rocks are heated and melt.

Some magma move sideways in opposite directions.Magma cools and sinks.Magma expands and rises at divergent boundaries.Magma expands and rises at divergent boundaries.

6 0
2 years ago
In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o
Kitty [74]

Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1}  )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w

The specific gravity of the wood chips is 0.494.

The average volume of a log is

V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2}  \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}

The weight of one log is

M=\rho*V=0.494*\rho_w*12.57  ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm

To provide 3617 ton/day of wood chips, we need

n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min

The feed rate of logs is 14.3 logs/min.

7 0
3 years ago
Read 2 more answers
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