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3 years ago

I WILL GIVE BRAINLEST PLS HELP

Chemistry

2 answers:

Elan Coil [88]3 years ago

4 0

Option C. H3PO23

Please give me the crown...

P.S this was in my friends question,so sgr got correct believe me .

Gekata [30.6K]3 years ago

3 0

Answer:I need it

Explanation: I have the same quision in my chemistry exam but i dont have the answer

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Calculate the mass of a body<br>Whose volume is<br>Is 2cm3 and<br> density is 520cm3

MrRissso [65]

Answer:

The answer is

Explanation:

Density = mass / volume

mass = density × volume

volume = 2cm³

density = 520g/cm³

mass = 2 × 520

= 1040g

Hope this helps you

6 0

3 years ago

A tank at is filled with of dinitrogen monoxide gas and of boron trifluoride gas. You can assume both gases behave as ideal gase

blsea [12.9K]

Answer:

(1). Mole fraction = 0.152 for sulfur tetrafluoride gas.

Mole fraction = 0.848 For dinitrogen monoxide gas.

(2). Partial Pressure for dinitrogen monoxide gas = 187 kPa

Partial Pressure for sulfur tetrafluoride gas = 33.4 kpa.

(3). Total Partial Pressure = 220.4 kpa.

Explanation:

So, we are given the following data or parameters or information in the question above;

• Volume of the tank = 5.00L per tank;

• Temperature of the tank = 7.03°C;

• The mass of the content in the tank =

17.7g of dinitrogen monoxide gas and

7.77g of sulfur tetrafluoride gas.

So, we will be making use of the formulae below to calculate the MOLE FRACTION:

Moles, n= mass/molar mass and mole fraction = n(1)/ n(1) + n(2) per each constituents.

Moles, n1 = 17.7g of dinitrogen monoxide gas/ 44 grams per mole. =0.4023 moles.

Moles, n2 = 7.77g of sulfur tetrafluoride gas/ 108.1 grams per mole. = 0.07188 moles.

Total numbers of moles = n1 + n2 = 0.47415 moles

Mole fraction =0.4023 / 0.47415 = 0.848 of dinitrogen monoxide gas.

Mole fraction = 0.07188/0.47415 = 0.152 of sulfur tetrafluoride gas.

PART TWO: CALCULATE THE PARTIAL PRESSURE AND TOTAL PRESSURE BY USING THE FORMULA BELOW;

pressure × volume = number of moles × gas constant, R × temperature.

Pressure = n × R × T/ V.

For dinitrogen monoxide gas. ;

Partial Pressure = 0.4023 × 8.314 × 280.03 / 5 × 10^-3 = 187 kPa.

For sulfur tetrafluoride gas

Partial Pressure = 0.07188 × 8.314 ( × 280.03 / 5 × 10^-3. = 33.4 kpa.

(3). Total pressure = (187 + 33.4)kpa = 220.4 kpa

6 0

3 years ago

An iron ore was found to contain 92.0% pyrite (fool's gold). Pyrite has the formula FeS2and is 46.5% iron. How many grams of iro

Liula [17]

Firstly calculate the grams in the last 8 percent before moving onto the pyrite section.
50.8x0.08=4.064g
We know that iron ore in this case has 92 percent pyrite which contains 46.5 percent iron so we do 50.8x0.92=46.736g from this we need to find 46.5 percent of the iron content in the 92 percent pyrite section then add this answer to the 8 percent of iron ore we found at the start.46.736x0.465=21.73224g
21.73224g+4.064=25.79624g of iron ore 25.8g(3sf)

4 0

4 years ago

People tend to speak more quietly in restaurants than they do when they are having an ordinary conversation Restaurant conversat

MArishka [77]

Answer:

A loud ordinary conversation following the supplied information in the question is about 4500 dB. But, in the official decibel system measure a loud conversation does not overcome 100 dB.

Explanation:

Using the supplied data of the exercise, we say that in a restaurant conversation the value is 45 dB. If we multiply this by 100 we will have a value for a laud ordinary conversation.

45×100 = 4500 dB.

but as I mentioned in the answer, in the official decibel system measure a loud conversation between 2 man reaches a maximal of 100 dB.

6 0

4 years ago

Will get Brainliest

Anestetic [448]

Hydrogen gas and oxygen gas react to form liquid water according to the following equation:

2H₂ + O₂ → 2H₂O

a. Converting our given masses of each gas to moles, we have:

(25 g H2)/(2 × 1.008 g/mol) = 12.4 mol H2; and

(25 g O2)/(2 × 15.999 g/mol) = 0.781 mol O2.

From the equation, two moles of H2 react with every one mole of O2. To fully react with 12.4 moles of H2, as we have here, one would need 6.2 moles of O2, which is far more than what we're actually given. Thus, the oxygen is our limiting reactant, and as such it will be the first reactant to run out.

b. Since O2 is our limiting reactant, we use it for determining how much product, in this case, H2O, is produced. From the equation, there is a 1:1 molar ratio between O2 and H2O. Thus, the number of moles of H2O produced will be the same as the number of moles of O2 that react: 0.781 moles of H2O. The mass of water produced would be (0.781 mol H2O)(18.015 g/mol) ≈ 14 grams of water (the answer is given to two significant figures).

c. Since the hydrogen reacts with the oxygen in a 2:1 ratio, twice the number of moles of oxygen in hydrogen is consumed: 0.781 mol O2 × 2 = 1.562 mol H2. Since we began with 12.4 moles of H2, the remaining amount of excess H2 would be 12.4 - 1.562 = 10.838 mol H2. The mass of the excess hydrogen reactant would thus be (10.838 mol H2)(2 × 1.008 g/mol) ≈ 22 grams of hydrogen gas (the answer is given to two significant figures).

3 0

3 years ago