Help help help help help help

How many gallons of concentrated sulfuric acid (density of 1.84 g/cm3) are required to have a mass of 184.33 pounds?

KengaRu [80]

After series of conversion from one unit to another, the number of gallons was found to be 12 litres

<h3>Density, Mass and Volume</h3>

Given Data

Density of Sulfuric acid = 1.84 g/cm3

Mass of Sulfuric acid = 184.33 Pounds

Conversion from pounds to Gram

1 kg ---------2.204 lb

x kg ---------184.33 lb

x = 184.33/2.204

x = 83.634 kg

hence the mass in gram = 83.634*1000

mass in gram = 83634 g

Now let us find the volume

We know that density = mass/volume

volume = mass/density

volume = 83634/1.84

volume = 45453.26 cm^3

Convert cm^3 to litres

Volume in litres = 45453.26 /1000

Volume in litres = 45.45 Litres

Convert from Litres to Gallons

1 gallon = 3.79 L

x gallons = 45.45 L

x = 45.45/3.79

x gallons = 11.99 L

Approx = 12 Litres

Learn more about density here:

brainly.com/question/1354972

7 0

2 years ago

What would it be like if Earth did not rotate?

kolezko [41]

There would be no day and night :(

7 0

3 years ago

Read 2 more answers

A solution is prepared by combining 5.00 mL of 4.8x10-4 M NaSCN solution, 2.00 mL of 0.21 M Fe(NO3)3 solution and 13.00 mL of 0.

Tcecarenko [31]

Answer:

The analytical concentrations of thiocyanate ions:

$[SCN^-]=0.00012 mol/L$

The analytical concentrations of ferric ions:

$[Fe^{3+}]=0.063 mol/L$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

1) Moles of sodium thiocyanate = n

Volume of sodium thiocyanate solution = 5.00 mL = 0.005 L

(1 mL = 0.001L)

Molarity of the sodium thiocyanate = $4.8\times 10^{-4} M$

$n=4.8\times 10^{-4} M\times 0.005 L=2.4\times 10^{-6}mol$

1 mole of sodium thiocyanate has 1 mol of thiocyante ions.

So, moles of thioscyanate ions in $2.4\times 10^{-6}mol$ of NaSCN.

$=1\times 2.4\times 10^{-6}mol=2.4\times 10^{-6}mol$

2) Moles of ferric nitrate = n'

Volume of ferric nitrate solution = 2.00 mL = 0.002 L

Molarity of the ferric nitrate = 0.21 M

$n'=0.002 M\times 0.21 L=0.00042 mol$

1 mole of ferric nitrate has 3 moles of ferric ions.

So number of moles of ferric ions in 0.00042 moles of ferric nitrate is :

$3\times 0.00042 mol=0.00126 mol$

Volume of nitric acid = 13.00 mL

Total volume by adding all three volumes of solutions = V

V = 5.00 mL + 2.00 mL + 13.00 mL = 20.00 mL = 0.020 L

The analytical concentrations of thiocyanate ions:

$[SCN^-]=\frac{2.4\times 10^{-6}mol}{0.020 L}=0.00012 mol/L$

The analytical concentrations of ferric ions:

$[Fe^{3+}]=\frac{0.00126 mol}{0.020 L}=0.063 mol/L$

3 0

3 years ago

Read 2 more answers

An unknown compound containing only C and H was burnt, yielding 10.2 g of CO2 and 6.3 g of H2O. With a molecular weight of about

Lisa [10]

Answer:

$C_2H_6$

Explanation:

Hello.

In this case, we can see that the mass of carbon of the unknown compound comes from the yielded mass of carbon dioxide, thus, we compute the moles of carbon as follows:

$m_C=10.2gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1mol C}{1molCO_2}=0.232 molC$

Moreover, the mass of hydrogen comes from the yielded water, therefore we can also compute the moles of water:

$m_H=6.3gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH_2}{1molH_2O} =0.7molH$

Then, to find the subscripts in the empirical formula, we divide by the moles of carbon as the smallest:

$C:\frac{0.232}{0.232}=1\\ \\H:\frac{0.7}{0.232}=3$

Whose molar mass is:

$M_{CH_3}=12+1*3=15g/mol$

Thus, the ratio of the molecular formula to the empirical formula is:

$\frac{30}{15}=2$

Therefore, the molecular formula is twice the empirical formula:

$C_2H_6$

Which is actually ethane.

Regards.

3 0

3 years ago

Calcular el valor de la presión osmótica que corresponde a una solución que contiene 2 moles de soluto en un litro de solución a

Brums [2.3K]

El valor de la presión osmótica que corresponde a una solución que contiene 2 moles de soluto en un litro de solución a una temperatura de 17°C es 47.56 atmósferas.

La presión osmótica se puede calcular usando la siguiente ecuación:

$\pi = \frac{nRT}{V}$ (1)