If a compound has two atoms of aluminum (AI) and three atoms of oxygen (O) what would its chemical formula look like?

What mass of Na2SO4is needed to make 2.5 L of 2.0 Msolution? (Na = 23 g; S = 32 g; O = 16 g)

ivanzaharov [21]

Answer:

mass (g) needed = 710.2 grams Na₂SO₄(s)

Explanation:

Needed is 2.5 Liters of 2.0M Na₂SO₄; formula wt Na₂SO₄ = 142.04g/mol.

mass (grams) of Na₂SO₄(s) = Molarity needed x Volume needed in Liters x Formula Wt of solute

mass (grams) of Na₂SO₄(s) = (2.5L)(2.0M)(142.04g/mol) = 710.2 grams Na₂SO₄(s)

Mixing: Transfer 710.4 grams Na₂SO₄ into mixing vessel and add water-solvent up to but not to exceed 2.5 Liters total volume. Mix until dissolved.

Gives 2.5 Liters of 2.0M Na₂SO₄(aq) solution.

5 0

3 years ago

Combustion of hydrocarbons such as butane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosp

Dahasolnce [82]

Answer:

1. C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

2. V = 596L

Explanation:

Butane (C₄H₁₀) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) thus:

C₄H₁₀ + O₂ → CO₂ + H₂O

1. The balanced chemical equation is:

C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

2. 0,360kg of butane are:

360g× $\frac{1mol}{58,12g}$ =6,19moles of butane

These moles of butane are:

6,19moles of butane× $\frac{4CO_2}{1molButane}$ = 24,8 moles CO₂

Using V=nRT/P

Where:

n are moles (24,8 moles CO₂); R is gas constant (0,082atmL/molK); T is temperature, 20°C (293,15K); and P is pressure (1atm).

Volume (V) is:

V = 596L

I hope it helps!

8 0

3 years ago

HELP! Answer the question below PLEASE

garri49 [273]

Answer: Francesco Redis Experiment part 1 : The maggots were not observed in setup B and setup C because the jars were covered.

part 2 : He discovered that maggots appeared on the meat in the control jar, the jar left open. No maggots appeared in the jar covered with gauze.

Pasteur Experiment

part 1:In the pastures Experiment the microorganisms come from the air.

part 2:Louis Pasteur's pasteurization experiment illustrates the fact that the spoilage of liquid was caused by particles in the air rather than the air itself.

Explanation:

8 0

3 years ago

What is the answer for this question?

andrew11 [14]

Answer:

See explanation, some might be graded as wrong if it's an automatic grading system but most

Explanation:

1. lose
2. valence
3. noble (Atoms don't actually always do this, but since the word gas is after the blank, it is the only option)

4. 4
5. have
6. 10

5 and 6 are a little ambiguous and could have many answers

5 0

1 year ago

10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of

Dimas [21]

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and $7.2m^3$ respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 10 kPa

$P_2$ = final pressure of gas = 15 kPa

$V_1$ = initial volume of gas = $10m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $50^oC=273+50=323K$

$T_2$ = final temperature of gas = $75^oC=273+75=348K$

Now put all the given values in the above equation, we get:

$\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}$

$V_2=7.2m^3$

The new volume of carbon dioxide gas is $7.2m^3$

Now we have to calculate the new density of carbon dioxide gas.

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

Formula for new density will be:

$\rho_2=\frac{P_2M}{RT_2}$

where,

$P_2$ = new pressure of gas = 15 kPa

$T_2$ = new temperature of gas = $75^oC=273+75=348K$

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

$\rho$ = new density

Now put all the given values in the above equation, we get:

$\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}$

$\rho_2=0.2281g/L$

The new density of carbon dioxide gas is 0.2281 g/L

5 0

3 years ago