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user100 [1]
3 years ago
6

Look at the spectrums of a star moving towards Earth and a motionless star.

Chemistry
2 answers:
Kay [80]3 years ago
8 0

The answer is; The spectrum of a star shifts towards the blue region when the star moves towards Earth.

This is referred to as the Doppler shift effect. As the star moves closer to the observer, the wavelength of light it produces is ‘compressed’ hence changing its relative frequency (just like the pitch of the siren of an ambulance approaching you).


AfilCa [17]3 years ago
6 0

it is Blue shifted when moving towards Earth

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3.8 g/mL

Explanation:

From the question given above, the following data were obtained:

Volume of Water = 60 mL

Volume of Water + Object = 73.5 mL

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Next, we shall determine the volume of the object. This can be obtained as follow:

Volume of Water = 60 mL

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5 0
3 years ago
A 25.00 mL sample of 0.320 M KOH is titrated with 0.750 M HBr at 25 °C.
Fudgin [204]

Answer:

a. pH = 13.50

b. pH = 13.15

Explanation:

Hello!

In this case, since the undergoing chemical reaction between KOH and HBr is:

HBr+KOH\rightarrow KBr+H_2O

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:

pOH=-log([OH^-])=-log(0.320)=0.50

Thus, the pH is:

pH+pOH=14\\pH=14-pOH=14-0.50\\pH=13.50

Which is the same answer for a and b as they ask the same.

Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

n_{KOH}=0.02500mL*0.320mol/L=0.00800mol\\\\n_{HBr}=0.005L*0.750mol/L=0.00375mol

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

n_{KOH}^{remaining}=0.00425mol

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):

[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M

So the pOH and the pH turn out:

pOH=-log(0.142)=0.849\\pH+pOH=14\\pH=14-pOH=14-0.849\\pH=13.15

Best regards!

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