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3 years ago

Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of

Chemistry

1 answer:

VashaNatasha [74]3 years ago

7 0

Answer:

$2al + 3h2so4 = > al2(so4)3 + 3h2 \\ eqvt \\ = > \frac{5.4}{27} = > \frac{5 \times 50}{1000} \\ .2 \: mol = > .25 \: mol \\ \\ so \: .25 \: is \: limiting \: reagent \\ according \: to \: ideal \: gas \: equation \\ pv = nrt \\ v = \frac{nrt}{p} \\ v = \frac{.25 \times .082 \times 300}{1} \\ = \frac{1}{4} \times 8.2 \times 3 \\ = 2.05 \times 3 \\ = 6.15litre \\ thank \: you$

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Assume that the complete combustion of one mole of glucose to carbon dioxide and water liberates 2870 kJ/mol2870 kJ/mol ( Δ????°

MaRussiya [10]

Answer:

Number of contraction cycles that could theoretically be fueled by the complete combustion of one mole of glucose is around 17

Explanation:

Energy released during the complete combustion of 1 mole glucose = 2870 kJ

Energy required/muscle contraction cycle = 67 kJ/contraction

Energy conversion efficiency = 39%

Actual amount of energy converted to contraction per mole of glucose is:

$=\frac{39}{100} *2870 kJ=1119.3 kJ$

Total contraction cycles fueled by the above energy is:

$=\frac{1119.3\ kJ}{67\ kJ/contraction} =16.7\ i.e.\ around\ 17\ contractions$

5 0

3 years ago

What would happen if a Continental Tropical Front meets a Maritime Polar front?

rewona [7]

Answer:

As soon as a front passes a particular location, the weather could change from a steamy maritime tropical air mass to a dry.

Explanation:

hope this helps

7 0

3 years ago

Read 2 more answers

ILL GIVE YOU BRAINLIEST!!!!Which of the following is an example of a description of climate? *

Murrr4er [49]

Answer:

The relative humidity in Raleigh, NC

You can search it up online. Search up what is climate change and it will give you "the weather conditions prevailing in an area in general or over a long period." (On bing)

Explanation:

Can I have brainliest? It would help me out, if not thanks anyways! Hope this helped and have a nice day!

6 0

3 years ago

The concentration of protein in a urine sample is calculated to be 2.77 μg/mL. What is the concentration of this solution in uni

maxonik [38]

Answer:

The concentration of this solution in units of pounds per gallon is $2.776*10^{-5} \frac{lb}{gal}$

Explanation:

Units of measurement are established models for measuring different quantities. The conversion of units is the transformation of a quantity, expressed in a certain unit of measure, into an equivalent one, which may or may not be of the same system of units.

In this case, the conversion of units is carried out knowing that 1 μg are equal to 2.205*10⁻⁹ Lb and 1 mL equals 0.00022 Gallons. So

$2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}$

If 1 μg equals 2.205*10⁻⁹ lb, 2.77 μg how many lb equals?

$lb=\frac{2.77ug*2.205*10^{-9}lb }{1ug}$

lb=6.10785*10⁻⁹

So, 2.77 μg= 6.10785*10⁻⁹ lb

Then:

$2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}=\frac{6.10785*10^{-9}lb }{mL} =\frac{6.10785*10^{-9}lb }{0.00022 gal} =\frac{6.10785*10^{-9}lb }{0.00022 gal}$

You get:

$2.77 \frac{ug}{mL} = 2.776*10^{-5} \frac{lb}{gal}$

The concentration of this solution in units of pounds per gallon is $2.776*10^{-5} \frac{lb}{gal}$

6 0

3 years ago

Consider the following data for air trapped in a flask: Pressure = 0.988 atm Room Temperature = 23.5°C Volume of the flask = 1.0

ddd [48]

Answer:

total mass will be = = 1.207g

Explanation:

First what is given

Pressure P= 0.988 atm Room TemperatureT = 23.5°C= 296.5 K

Volume V= 1.042 L

Nitrogen in air is 80 % (moles number) = 0.8

Ideal gas constant R = 0.0821 L atmmol-1K-1

Given mass of N = 14.01 g/mol

Given mass of oxygen O = 16.00 g/mol

Total number of moles = ?

So first we have to find the total number of moles by using formula

Total number of moles

n = PV/ RT

adding the values

moles n= 0.988atm x 1.042L / (0.0821L-atm/mole-K x 296.6K)

= 1.0294 / 24.35

= 0.042 moles (total number of moles)

So by using Nitrogen percentage

Moles of nitrogen = total moles x 80/100

= 0.042moles x 0.8

So moles of O2= Total moles – moles of N2

= 0.042moles - 0.034moles

Moles of O2 = 0.008moles

Now for finding the mass of the N2 and oxygen

Mass of Nitrogen N2 = no of moles x molar mass

= 0.034 x 28

= 0.952 g

Mass of oxygen O2 = no of moles x molar mass

= 0.008 x 32

= 0.256 g

total mass will be = Mass of Nitrogen N2 + Mass of oxygen O2

=0.952 g + 0.256 g

=1.207g

6 0

3 years ago