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2 years ago

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I need to right a journal entry for school but i don't know what to put Write a journal entry from the perspective of a particle

of light,
called a photon. Your story should follow the photon as it
travels from a star in the Milky Way that is 25 light years away
from Earth, all the way to the eye of a young person gazing up
at the night sky through their bedroom window. Compose your
entry in complete sentences.

it is for my 5 grad sicenc

1 answer:

polet [3.4K]2 years ago

5 0

Answer:

here you go

Explanation:

someone help me on this but i still got a hundred

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An emulsifying agent is typically characterized by having ____. a. one polar end c. two polar ends b. one nonpolar end d. one po

tangare [24]

Answer:

An emulsifying agent is typically characterized by having <u><em>d. one polar end and one nonpolar end.</em></u>

Explanation:

Emulsifiers are substances that have the ability to bind, for example, fats with those substances that have mostly water in their conformation. In other words, the emulsifier facilitates mixtures of two or more immiscible liquid substances.

This is because the molecules of an emulsifier are often lipophilic (attract oil) at one end and hydrophilic (attract water) at the other. In other words it consists of a polar (hydrophilic) head group and a non-polar (hydrophobic) tail.

<u><em>An emulsifying agent is typically characterized by having d. one polar end and one nonpolar end.</em></u>

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3 years ago

Conservation of mass to balance the following reaction. Na2S+ KI= NaI+ K2S

lesantik [10]

Answer:

1+2=2+1 is the answer for the question

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3 years ago

How much space do 2x10 22 copper pennies take up

soldi70 [24.7K]

Answer:

40+22=

Explanation:

62

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3 years ago

Calculate the number of moles and the mass of the solute in each of the following solutions:

frosja888 [35]

<u>Answer:</u>

<u>For a:</u> The number of moles of KI are $2.7\times 10^{-5}$ and mass is $4.482\times 10^{-3}g$

<u>For b:</u> The number of moles of sulfuric acid are $1.65\times 10^{-5}$ and mass is $1.617\times 10^{-3}g$

<u>For c:</u> The number of moles of potassium chromate are $2.84\times 10^{-2}$ and mass is 5.51 g.

<u>For d:</u> The number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

To calculate the number of moles of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(2)

<u>For a:</u>

Molarity of KI = $8.23\times 10^{-5}M$

Volume of solution = 325 mL = 0.325 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$8.25\times 10^{-5}mol/L=\frac{\text{Moles of KI}}{0.325L}\\\\\text{Moles of KI}=2.7\times 10^{-5}mol$

Now, using equation 2, we get:

Moles of KI = $2.7\times 10^{-5}mol$

Molar mass of KI = 166 g/mol

Putting values in equation 2, we get:

$2.7\times 10^{-5}mol=\frac{\text{Mass of KI}}{166g/mol}\\\\\text{Mass of KI}=4.482\times 10^{-3}g$

Hence, the number of moles of KI are $2.7\times 10^{-5}$ and mass is $4.482\times 10^{-3}g$

<u>For b:</u>

Molarity of sulfuric acid = $22\times 10^{-5}M$

Volume of solution = 75 mL = 0.075 L

Putting values in equation 1, we get:

$22\times 10^{-5}mol/L=\frac{\text{Moles of sulfuric acid}}{0.075L}\\\\\text{Moles of }H_2SO_4=1.65\times 10^{-5}mol$

Now, using equation 2, we get:

Moles of sulfuric acid = $1.65\times 10^{-5}mol$

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 2, we get:

$1.65\times 10^{-5}mol=\frac{\text{Mass of }H_2SO_4}{98g/mol}\\\\\text{Mass of }H_2SO_4=1.617\times 10^{-3}g$

Hence, the number of moles of sulfuric acid are $1.65\times 10^{-5}$ and mass is $1.617\times 10^{-3}g$

<u>For c:</u>

Molarity of potassium chromate = $0.1135M$

Volume of solution = 0.250 L

Putting values in equation 1, we get:

$0.1135mol/L=\frac{\text{Moles of }K_2CrO_4}{0.250L}\\\\\text{Moles of }K_2CrO_4=2.84\times 10^{-2}mol$

Now, using equation 2, we get:

Moles of potassium chromate = $2.84\times 10^{-2}mol$

Molar mass of potassium chromate = 194.2 g/mol

Putting values in equation 2, we get:

$2.84\times 10^{-2}mol=\frac{\text{Mass of }K_2CrO_4}{194.2g/mol}\\\\\text{Mass of }K_2CrO_4=5.51g$

Hence, the number of moles of potassium chromate are $2.84\times 10^{-2}$ and mass is 5.51 g.

<u>For d:</u>

Molarity of ammonium sulfate = 3.716 M

Volume of solution = 10.5 L

Putting values in equation 1, we get:

$3.716mol/L=\frac{\text{Moles of }(NH_4)_2SO_4}{10.5L}\\\\\text{Moles of }(NH_4)_2SO_4=39.018mol$

Now, using equation 2, we get:

Moles of ammonium sulfate = 39.018 mol

Molar mass of ammonium sulfate = 132.14 g/mol

Putting values in equation 2, we get:

$39.018mol=\frac{\text{Mass of }(NH_4)_2SO_4}{132.14g/mol}\\\\\text{Mass of }(NH_4)_2SO_4=5155.84g$

Hence, the number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

3 0

3 years ago

stomach acid contains dilute hydrochloric acid write a chemical equation to show the neutralization of stomach acid by milk of m

Eddi Din [679]

Answer:

Mg(OH)2(s) + 2HCl(aq) → 2H2O(l) + MgCl2(aq)

Explanation:

4 0

2 years ago