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andreyandreev [35.5K]
3 years ago
7

Single covalent bonds are also referred to as ___________.

Chemistry
2 answers:
liubo4ka [24]3 years ago
8 0

Answer:

Single covalent bond are also referred to as sigma bonds

iren2701 [21]3 years ago
3 0
A I think I hope it’s right! Good luck ;)
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How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.88 ∘c?
e-lub [12.9K]
The freezing point depression is calculated through the equation,
                                    ΔT = (kf)  x m 
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
                                   5.88 = (1.86)(m)
m is equal to 3.16m

Recall that molality is calculated through the equation,
                                  molality = number of mols / kg of solvent
                                       number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.

6 0
3 years ago
? Match the states of matter to their properties. Drag the items on the left to the correct location on the right. solids Indefi
Fantom [35]

Answer:

Solids: definite shape and definite volume (highest density)

Liquid: indefinite shape and definite volume (glide past each other)

Gas: indefinite shape and indefinite volume (lowest density)

Explanation:

look at the answer

8 0
3 years ago
A 8L sample of gas at 250 K is cooled to 75 K. What is the volume of the gas after it is cooled?
Stells [14]

by \: Charles'\: law: \:  \\  \frac{v_{1}}{t_{1}}  =  \frac{v_{2}}{t_{2}}  \\

Where v is the volume(in L) and t is the temperature(in °K)

\frac{8}{250}  =  \frac{v_{2}}{75} \\\\4/5 =v_2/3 \\\\\boxed{v_{2} = \frac{12}{5}}\\\\\\\huge{=2.5 l}

7 0
2 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0
3 years ago
What is the correct order of energy transformation in a windmill?
Vika [28.1K]

Answer:

A wind turbine transforms the mechanical energy of wind into electrical energy. A turbine takes the kinetic energy of a moving fluid, air in this case, and converts it to a rotary motion.

hope it helps (^^)

# Cary on learning

8 0
2 years ago
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