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3 years ago

Which of the following is not true of a scientific theory?

Chemistry

1 answer:

Furkat [3]3 years ago

3 0

Answer: a,b

Explanation: because a scientific theory is something that been thought of and tested multiple times.

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Consider the combustion of octane (C8H18)

sesenic [268]

The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
mass of CO₂ formed is 591.8 g

3 0

3 years ago

Nault 25000L 250 our = 25000L 15040 8. How many grams of CaCl2 are needed to make 150.0 mL of a 0.500 M CF solution? (Note: CaCl

yuradex [85]

Answer:

You need 8,324 g of CaCl₂ yo make this solution

Explanation:

Molarity is a way to express concentration in a solution, in units of moles of solute per liter of solution.

To know the grams of CaCl₂ it is necessary to know, first, the moles of this substance with the desired volume and concentration , thus:

0,1500 L × $\frac{0,500 mol}{L}$ = 0,075 CaCl₂ moles

Now, with the molar mass of CaCl₂ you will obtain the necessary grams, thus:

0,075 CaCl₂ moles × $\frac{110,98 g}{mol}$ = 8,324 g of CaCl₂

So, you need 8,324 g of CaCl₂ to make 150,0 mL of a 0,500M solution

I hope it helps!

3 0

3 years ago

When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem

masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

Step 1: Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

Step 2: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

Step 3: Define the formula

Pressure is constant. → the heat evolved from the reaction is equivalent to the enthalpy of reaction.

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for water (1g/mL and 4.184 J/g°C)

7 0

3 years ago

What is the mass in grams of 0.7350 moles of sodium?

Alik [6]

Answer:

Explanation:

1 mol of sodium = 23 grams (use the number on your periodic table).

0.7350 mol sodium = x

Cross multiply

1*x = 0.7350 * 23

x = 16.905

You will get slightly less than this, depending on your periodic table. But the method will be the same.

4 0

3 years ago

What is the volume of 0.20 moles of helium at STP?

stealth61 [152]

The volume of 0.20 moles of helium at STP is 4.5 liters.

Explanation:

Given:

Number of moles = 0.20 moles

To Find:

The volume of Helium at STP =?

Solution:

According to ideal gas law

PV = nRT

where

P is pressure,

V is volume,

n is the number of moles

R is the gas constant, and

T is temperature in Kelvin.

The question already gives us the values for p and T ,because helium is at STP. This means that temperature is 273.15 K and pressure is 1 atm .

We also already know the gas constant. In our case, we'll use the value of

0.08206 L atm/K mol since these units fit the units of our given values the best

On substituting these values we get

$1 atm \times V = 0.20 moles \times 0.08206 L atm/K mol$

$V = \frac{0.20 moles \times 0.08206 L atm/K mol }{ 1 atm}$

$V = \frac{4.482}{1}$

V = 4.5 Liters

7 0

3 years ago