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Savatey [412]
3 years ago
15

What are four indications that a chemical reaction bay be taking place

Chemistry
1 answer:
Mariana [72]3 years ago
3 0
When a chemical reaction takes place it changes the composition of the reactants. The ways to tell if a reaction is occurring are,
1-Release of heat
2-Production of a gas
3-Formation of a precipitate 
4-Change in color
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I have no idea how to do this
alexandr402 [8]
<span>It is known that amu is equal to the molar mass of 1 mole of substance. Therefore,
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Weight of one mole of uranium = 238 g/mol


The molar mass or molecular weight for U205 is 556 g/mol.
(238 x 2) + (16 x 5) = 556 g/mol
7 0
4 years ago
Read 2 more answers
Calculate the cell potential, Ecell, of a concentration cell containing a 0.1 M solution of Zn2 and a 1.0 M solution of Zn2
Agata [3.3K]

Answer:

The cell potential is 0.0296V

Explanation:

Please see the attachment below

3 0
3 years ago
Help pleaseeee. over dilutions in chemistry
atroni [7]

Answer:

5 L

Explanation:

We'll begin by calculating the molarity of the CaCl₂ solution. This can be obtained as follow:

Mole of CaCl₂ = 0.5 mole

Volume = 2 L

Molarity =?

Molarity = mole /Volume

Molarity = 0.5 / 2

Molarity = 0.25 M

Finally, we shall determine the volume of the diluted solution. This can be obtained as follow:

Molarity of stock solution (M₁) = 0.25 M

Volume of stock solution (V₁) = 2 L

Molarity of diluted solution (M₂) = 0.1 M

Volume of diluted solution (V₂) =?

M₁V₁ = M₂V₂

0.25 × 2 = 0.1 × V₂

0.5 = 0.1 × V₂

Divide both side by 0.1

V₂ = 0.5 / 0.1

V₂ = 5 L

Thus the volume of the diluted solution is 5 L

5 0
3 years ago
Which of the following is true about photosynthesis?
Marat540 [252]
The correct answer is A. Photosynthesis is an exothermic reaction. The reaction releases energy. For the second question, the correct answer is A. Atomic theory is subject to change if new information is discovered. Lastly, the correct answer for the third question is option A. <span>Two nonmetals from Group 16 are likely to form a metallic bond.</span>
4 0
4 years ago
Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains
erastova [34]

Given that:

  • The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
  • normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K  = 272.6 K
  • volume  = 250 mL = 0.250 L
  • Mass of butane = 0.8 g
  • the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}

where;

P1 and P2 correspond to the temperature at T1 and T2.

∴

replacing the values into the given equation, we have;

\mathbf{In \dfrac{P_2}{1\  atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}

\mathbf{In \dfrac{P_2}{1\  atm} =-(0.852053785)}

\mathbf{P_2=0.427 \ atm}

As such, at -22° C; the vapor pressure = 0.427 atm

Now, using the ideal gas equation:

PV = nRT

where:

  • P = Pressure
  • V = volume
  • n = number of moles of butane
  • R = universal gas constant
  • T = temperature

∴

Making (n) the subject of the formula:

\mathbf{n = \dfrac{PV}{RT}}

\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}

\mathbf{n =0.00518 mol}

We all know that the standard molecular weight of butane = 58.12 g/mol

∴

Using the relation for the number of moles which is:

\mathbf{number \  of \  moles = \dfrac{mass}{molar mass}}

mass = 0.00518 mole × 58.12 g/mol

mass = 0.301 g

∴

The mass of butane in the flask = 0.301 g

But the mass of the butane present as a liquid in the flask is

= 0.8 g - 0.301 g

= 0.499 g

In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g

Learn more about vapourization here:

brainly.com/question/17039550?referrer=searchResults

7 0
3 years ago
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