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lions [1.4K]
3 years ago
10

ASAP!!!!PLS I NEED IT FAST ILL MARK BRAINLIEST!!!! POR FAVOR

Chemistry
1 answer:
Nataliya [291]3 years ago
6 0
The second one ......
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Identical rock types, identical fossils, and very similar mountain ranges are found on different continents that are separated b
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Answer:

These continents were joined together at one time in the past.

Explanation:

Identical rock types, identical fossils, and very similar mountain ranges are found on different continents that are separated by a wide ocean. Which of these BEST explains these observations?

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Fluids that have a lower viscosity than water
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Syrup, molasses, and honey have a lower viscosity than water

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One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silv
Flauer [41]

Answer:

1.1 × 10⁻⁴ M

Explanation:

Let's consider the following double displacement reaction.

CuCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s)+ Cu(NO₃)₂(aq)

We can establish the following relations:

  • The molar mass of AgCl is 143.32 g/mol.
  • The molar ratio of AgCl to CuCl₂ is 2:1

The moles of CuCl₂ that reacted to produce 7.7 mg of AgCl are:

7.7 \times 10^{-3} gAgCl.\frac{1molAgCl}{143.32gAgCl} .\frac{1molCuCl_{2}}{2molAgCl} =2.7 \times 10^{-5}molCuCl_{2}

The molarity of CuCl₂ is:

M=\frac{2.7 \times 10^{-5}molCuCl_{2}}{0.250L} =1.1\times 10^{-4} M

7 0
3 years ago
State the Constituents of the following alloys:magnalium and bronze​
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Magnalium: Magnesium and Aluminum

Bronze: Copper, Tin, Arsenic, Phosphorus, Aluminum, Manganese and Silicon (whichever you learned in class from those)
8 0
3 years ago
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i
Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane  ⇄  Isobutane

At equilibrium

1.0 M               2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M               -

After equilibrium reestablishes:

(1.50-x)M            (2.5+x)

The equilibrium expression will wriiten as:

K_c=\frac{[Isobutane]}{[Butane]}

2.5=\frac{2.5+x}{(1.50-x)}

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0
3 years ago
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