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Inessa [10]
3 years ago
7

I NEED ANSWER ASAP PLS (picture included)

Chemistry
1 answer:
mojhsa [17]3 years ago
7 0

Answer: Heres a vid talking about the topic and answer (it helpes!)

Explanation:

h t t p s : / / w w w . y o u t u b e . c o m / w a t c h ? v = i i k 2 5 w q I u F o

remove the spaces bc you know brainly -_-

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Suppose that the physical properties of two stars are identical in every way --- for example, same distance, same mass, same tem
larisa [96]

Answer:

By measuring it's Radical velocity using Doppler Phenomenon...

Explanation:

It is done by measuring the absorption spectrum produces by a star as a result of measuring Doppler's relative wavelength. The star with the lower speed must have lower absorption spectrum extent as compared to the faster one.

4 0
3 years ago
The reaction of an alkene and water in the presence of an acid catalyst to produce an alcohol is called ________.
pochemuha
The answer I believe is Hydration
5 0
2 years ago
Is solid sodium + liquid water to dissolved hydroxide + gaseous hydroxide a single replacement reaction
saul85 [17]
2Na + 2H20 ----> 2NaOH + H2
5 0
3 years ago
Magnesium element has three isotopes: 24Mg, 25Mg, and 26Mg. The natural abundance of the isotopes on Earth is 78.99% 24Mg with a
Dafna11 [192]

Answer:

24.305 to the nearest thousandth.

Explanation:

That is 0.7899*23.985 + 0.10* 24.985+0.1101*25.9825

= 24.305.

3 0
3 years ago
You placed 6.35 g of a mixture containing unknown amounts of BaO(s) and MgO(s) in a 3.50-L flask containing CO₂(g) at 30.0°C and
adelina 88 [10]

Mass of BaO in  initial mixture = 3.50g

Explanation:

Let mass of BaO in mixture be x g

mass of MgO in mixture be (6.35 - x) g

Initially CO_2

Volume = 3.50 L

Temp = 303 K

Pressure = 750 torr = 750 / 760 atm

Applying ideal gas equation

PV = nRT

n = PV / RT

(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303

(n)_CO_2 = 0.139 mole

Finally; mole of CO_2

n= PV /RT

((245/760) *3.5) / 303* 0.0821

(n)_CO_2 = 0.045 mole

Mole of CO_2 reacted = 0.139 - 0.045

=0.044 mole

BaO + CO_2  BaCO_3

Mgo + CO_2  MgCO_3

moles of CO_2 reacted = ( moles of BaO + moles of MgO)

moles of BaO in mixture = x / 153 mole

moles of MgO in mixture = 6.35 - x mole / 40

Equating,

x/ 153 +6.35/40 = 0.094

= x/153 + 6.35 / 40 - x/40 =0.094

= x (1/40 - 1153) = (6.35/40 - 0.094)

= x * 10.018464

= 0.06475

mass of BaO in mixture = 3.50g

5 0
2 years ago
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