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Scrat [10]
3 years ago
5

Describe how greater distance affects the appearance of an object’s size and brightness

Chemistry
1 answer:
jeka943 years ago
4 0

Answer:

Distance also affects the apparent brightness of an object. A distant, but very energetic star can appear dimmer to us than a lower-energy, but closer one. ... Each of their effective temperatures results in a different brightness for the object. The quasar, however, is very far away, and so appears dim.

Explanation:

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determine the concentration of 35.0mL of an acid solution that requires 25.1mL of a 0.50M NaOH to neutralize it
Degger [83]
Hello!

The general chemical equation for the neutralization of an acid HA with NaOH is the following:

HA(aq) + NaOH(aq) → NaA(aq) + H₂O(l)

For determining the concentration of the acid solution, we can use the equation shown below:

moles HA=moles NaOH \\  \\ cHA*VHA=cNaOH*VNaOH \\  \\ cHA= \frac{cNaOH*vNaOH}{VHA}= \frac{0,5M*25,1 mL}{35 mL}=0,36 M

So, the concentration of the Acid is 0,36 M

Have a nice day!
7 0
4 years ago
1)According to Newton's second law i the net force increases, the acceleration of the body
MissTica
The answer is B) Increases

The net force of an object is equal to the product of its acceleration and its mass
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Is it possible for a child to have a dna band that is not found in the mother's dna? Explain
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Nah ur adopted lol srry
6 0
3 years ago
Read 2 more answers
Carbon dioxide table
satela [25.4K]

Answer:

The table refers to emissions of CO2 from burning oil, coal and gas for energy use. Carbon dioxide also enters the Atmosphere from burning wood and waste materials and from some industrial processes such as cement production.

Explanation:

3 0
3 years ago
Be sure to answer all parts. The balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g) → 2Al
Arada [10]

Answer: a) Cl_2 is the limiting reagent

b)  0.27 g of AlCl_3 will be produced.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{0.80g}{27g/mol}=0.030moles

\text{Moles of} Cl_2=\frac{0.23g}{71g/mol}=0.003moles

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)

According to stoichiometry :

3 moles of Cl_2 require = 2 moles of Al

Thus 0.003 moles of Cl_2 will require=\frac{2}{3}\times 0.003=0.002moles  of Al

Thus Cl_2 is the limiting reagent as it limits the formation of product and Al is the excess reagent.

b) As 3 moles of Cl_2 give = 2 moles of AlCl_3

Thus 0.003 moles of Cl_2 give =\frac{2}{3}\times 0.003=0.002moles  of AlCl_3

Mass of AlCl_3=moles\times {\text {Molar mass}}=0.002moles\times 133g/mol=0.27g

Thus 0.27 g of AlCl_3 will be produced.

8 0
3 years ago
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