True or False: Some new technologies were created based on designs found in nature.

At 35.0°c and 3.00 atm pressure, a gas has a volume of 1.40 l. what pressure does the gas have at 0.00°c and a volume of 0.950 l

Leona [35]

Answer : The pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

Solution :

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 = 3 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 1.40 L

$V_2$ = final volume of gas = 0.950 L

$T_1$ = initial temperature of gas = $35^oC=273+35=308K$

$T_2$ = final temperature of gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{3atm\times 1.40L}{308K}=\frac{P_2\times 0.950L}{273K}$

$P_2=3.918atm$

Therefore, the pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

4 0

3 years ago

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A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)

elixir [45]

Answer: $\Delta H^{0}=-173.72$ kJ/mol

Explanation: Enthalpy Change is the amount of energy in a reaction - absorption or release - at a constant pressure. So, Standard Enthalpy of Formation is how much energy is necessary to form a substance.

The standard enthalpy of formation of HCl is calculated as:

$\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}$

$Ca(OH)_{2}_{(aq)}+2HCl_{(aq)}$ → $CaCl_{2}_{(s)}+2H_{2}O_{(l)}$

Standard Enthalpy of formation for the other compounds are:

Calcium Hydroxide: $\Delta H^{0}=$ -1002.82 kJ/mol

Calcium chloride: $\Delta H^{0}=$ -795.8 kJ/mol

Water: $\Delta H^{0}=$ -285.83 kJ/mol

Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.

Calculating:

$-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]$

$-17.2=-1367.46+1002.82-2\Delta H$

$2\Delta H=17.2-364.64$

$\Delta H=-173.72$

So, the standard enthalpy of formation of HCl is -173.72 kJ/mol

8 0

2 years ago

2. Nitric oxide contains 46.66% nitrogen and 53.34% oxygen. Water contains 11.21% hydrogen and 88.79% oxygen. Ammonia contains 1

Mrrafil [7]

Answer:

The law of reciprocal proportions states that if two elements react individually with a given weight of a third element, the ratio of the masses with which they combine with the third element are either the same or a simple multiple of the ratio of the masses with which they combine with each other

The compounds formed includes;

1) Nitric oxide, NO

Nitrogen = 46.66% × 30.01 = 14

Oxygen = 53.34% × 30.01 = 16

2) Water, H₂O

Hydrogen = 11.21% × 18.01528 = 2

Oxygen = 88.79% × 18.01528 ≈ 16

3) Ammonia, NH₃

Hydrogen = 17.78% × 17.031 ≈ 3

Nitrogen = 82.22% × 17.031 ≈ 14

The ratio of nitrogen to oxygen in nitric oxide = 14:16 = 7:8

The ratio of nitrogen to hydrogen in ammonia = 14:3

The ratio in which hydrogen and oxygen combine with nitrogen = 3/16

The ratio of hydrogen and oxygen combine with each other in water = 2/16

Therefore, the ratio with which hydrogen and oxygen combine with nitrogen, is (2/3) times the ratio with which they combine with each other, which verifies the law of reciprocal proportions

Explanation:

4 0

3 years ago

Fossil A is found in a rock layer above a layer containing Fossil B. Which fossil is probably older?

ch4aika [34]

Answer:

Fossil B

Explanation:

Fossil B is older because it's later is at the bottom

4 0

2 years ago

Hydrazine (N2H4) is used as rocket fuel. It reacts with oxygen to form nitrogen and water.

Marina86 [1]

Answer:

See explanation below for answers

Explanation:

This is a stochiometry reaction. LEt's write the overall reaction again:

N₂H₄ + O₂ ---------> N₂ + 2H₂O

This reaction is taking place at Standard temperature and pressure conditions (STP) which are P = 1 atm and T = 273 K. To know the volume of N₂ formed, we need to know first how many moles are formed, and this can be calculated with the reagents and the limiting reagent. Let's calculate the moles first of the reagents:

MM N₂H₄ = 32 g/mol; MM O₂ = 32 g/mol

mol N₂H₄ = 2000 / 32 = 62.5 moles

mol O₂ ? 2100 / 32 = 65.63 moles

Now that we have the moles, we need to apply the stochiometry and calculate the limiting reagent. According to the overall reaction we have a mole ratio of 1:1 between N₂H₄ and O₂, therefore:

1 mole N₂H₄ ---------> 1 mole O₂

62.5 moles ----------> X

X = 62.5 moles of O₂

But we have 65.63 moles, therefore, the limiting reactant is the N₂H₄.

We also have a 1:1 mole ratio with the N₂, so:

moles N₂H₄ = moles N₂ = 62.5 moles

Now that we have the moles, we can calculate the volume with the ideal gas equation:

PV = nRT

V = nRT / P

R: gas constant (0.082 L atm / K mol)

Replacing we have:

v = 62.5 * 0.082 * 273 / 1

V = 1399.13 L of N₂

Now, how many grams of the excess remains?, we know how many moles are reacting so, let's see how much is left:

moles remaining = 65.63 - 62.5 = 3.12 moles

then the mass of oxygen:

m = 3.12 * 32 = 100.16 g of O₂

7 0

3 years ago

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