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Alex777 [14]
3 years ago
9

Which term defines the distance from the focal point to either a lens or a mirror

Chemistry
1 answer:
nikklg [1K]3 years ago
3 0
The focal length, like you said it's the distance between the FOCAL point and the mirror.


Hope this helps.... :)
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Which of the following pairs of elements could possibly be in the same group? X has a 1+ ion; Y has a 1- ion. X tends to form a
exis [7]
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
5 0
3 years ago
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Based on the nebular Theory, which of these is a step in the formation of a new planetary system?
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A. evaporation of water
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2 years ago
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Salt in crude oil must be removed before the oil undergoes processing in a refinery. The
irina1246 [14]

Answer:

\large \boxed{0.64 \, \%}

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}

2. Calculate the mass of salt in the salty oil

\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}

3. Calculate the mass of salt in the spent water

\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g  

5. Concentration of salt in washed oil

\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}

3 0
3 years ago
The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4 .a. Write the chemical equation for the equilibrium that correspond
kvv77 [185]

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

   ΔGº= - RT ln Ka

   ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

  ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

6 0
3 years ago
the density of glycerin is 1.26 g/cm3. How many pounds/foot3 is this ? use the conversion rates of 454g/1 pound and 28,317cm3/1f
lyudmila [28]

Answer:

78.6 lb/ft³  

Step-by-step explanation:

Let's do this in steps.

1. Convert grams to pounds

D = (1.26 g/1 cm³) × (1 lb/454 g)

   = 2.775 × 10⁻³ lb/cm³

2. Convert cubic centimetres to cubic feet

D = (2.775 × 10⁻³ lb/1 cm³) × (28 317cm³/1 ft³)

   = 78.6 lb/ft³

6 0
2 years ago
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