What is the name for SO2

A 283.3-g sample of X2(g) has a volume of 30 L at 3.2 atm and 27 'C. What is element X?

professor190 [17]

we find the number of moles with the formula

pV=nRT

27C=300k

3.2*30=n*0.082*300

n=96/24.6=3.9 moles

we have the mass ant the number of moles so we can calculate the molar mass

283.3/3.9=72.64g/moles

that means X's mass is 72.64/2=36.3 so the element is Kr(Kriptonium)

4 0

2 years ago

What two quantities must be known to calculate the density of a sample of matter?

Tanya [424]

The answer is A, mass and volume.

The equation for density, <span>ρ,</span> is ρ = m/V. m is mass and V is volume.

7 0

3 years ago

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Many people in South America are subsistence farmers. Subsistence farming is...

galben [10]

growing only enough crops for food to fed the family

Explanation:

subsistence farming is the practice of growing crops only for own use .. some developing countries are only doing farming for their personal use so they are in the stage of developing ..in a country like nepal we not being able to trade our farming .if we want to export then also we arenot getting the good market ..

8 0

3 years ago

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The decomposition of NI3 to form N2 and I2 releases −290.0 kJ of energy. The reaction can be represented as 2NI3(s)→N2(g)+3I2(g)

EastWind [94]

Answer:

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

Explanation:

Mass of nitrogen triiodide = 20.0 g

Moles of nitrogen triiodide = $\frac{20.0 g}{395 g/mol}=0.05063 mol$

$2NI_3(s)\rightarrow N_2(g)+3I_2(g), \Delta H_{rxn}=-290.0 kJ$

According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

$\frac{-290.0 kJ}{2}\times 0.05063=-7.34 kJ$

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

3 0

3 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

2 years ago