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sweet [91]
2 years ago
7

Oil is a fossil fuel used to run cars, heat up homes, and produce electricity. Oil can be removed from the bottom of the ocean t

hrough drilling. Drilling machines dig deep down into the Earth. When oil is found, pipes carry it to the surface. Sometimes, accidents happen that cause the oil to spill into the oceans. The oil can not only kill marine organisms, but when it reaches the surface of the water, some chemicals evaporate, become part of the atmosphere, and pollute the air.
Based on the passage, which resources and organisms are affected by oil spills? Select three options.
air
land
water
birds in the air
trees on land
organisms in water HELP
Chemistry
2 answers:
Crazy boy [7]2 years ago
8 0
The answers would be organisms in water, water, and air.
KonstantinChe [14]2 years ago
7 0

Answer:

air, water and organisms in water

Explanation:

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What is the molarity of a solution of 14.0 g NH4Br in enough H2O to make 150 mL of solution?
kiruha [24]
The molarity of a solution equals to the mole number of the solute/the volume of the solution. For NH4Br, we know that the mole mass is 98. So the molarity is (14/98) mol /0.15 L=0.95 mol/L.
8 0
2 years ago
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A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of
allochka39001 [22]

Answer:

The answer to your question is   C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

                       44 g of CO₂ ---------------  12 g of C

                        52.8 g          ---------------  x

                        x = (52.8 x 12)/44

                        x = 633.6/44

                        x = 14.4 g of C

                        12 g of C ------------------  1 mol

                        14.4 g of C ---------------   x

                         x = (14.4 x 1)/(12)

                         x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

                         18 g of H₂O ---------------  2 g of H

                         21.6 g of H₂O -------------  x

                          x = (21.6 x 2) / 18

                         x = 2.4 g of H

                         1 g of H -------------------- 1 mol of H

                         2.4 g of H -----------------  x

                          x = (2.4 x 1)/1

                          x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

                           = 6.4 g

                         16 g of O ----------------  1 mol

                          6.4 g of O --------------  x

                          x = (6.4 x 1)/16

                          x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

                                C₃H₆O

8 0
2 years ago
A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is AgCl (s) + e− → Ag (s) + Cl
ANTONII [103]

Answer : The cell emf for this cell is 0.118 V

Solution :

The half-cell reaction is:

AgCl(s)+e^\rightarrow Ag(s)+Cl^-(aq)

In this case, the cathode and anode both are same. So, E^o_{cell} is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 1

E_{cell} = ?

[Cl^{-}{diluted}] = 0.0222 M

[Cl^{-}{concentrated}] = 2.22 M

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

E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}

E_{cell}=0.118V

Therefore, the cell emf for this cell is 0.118 V

4 0
2 years ago
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likoan [24]

Answer:

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Explanation:

6 0
2 years ago
The dipole moment (μ) of HBr (a polar covalent molecule) is 0.838D (debye), and its percent ionic character is 12.4 % . Estimate
myrzilka [38]

<span>When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, a dipole is established. The size of a dipole is measured by its dipole moment (</span>μμ). Dipole moment is measured in Debye units, which is equal to the distance between the charges multiplied by the charge (1 Debye equals 3.34×10−30Cm3.34×10−30Cm). The dipole moment of a molecule can be calculated by Equation 1.11.1:

μ = qr

where

<span> <span>μ⃗ μ→ is the dipole moment vector</span> <span>qiqi is the magnitude of the ithith charge, and</span> <span>r⃗ ir→i is the vector representing the position of ithith charge.</span> </span>

 

r = μ/q

<span>r = [0.838D(3.34×10−30 C⋅m/ 1D)]/ (1.6×10−19 C) *0.124
</span> r = 1.41 x10^-10 m

 

7 0
2 years ago
Read 2 more answers
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