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SCORPION-xisa [38]
3 years ago
7

We can identify unknown elements in a star by matching the ___________ spectrum of the star to those of known elements. *

Chemistry
1 answer:
const2013 [10]3 years ago
4 0
Emission spectrum, mass spectrum does not make sense and neither does atoms spectrum. You use the emission spectrum to compared known elements in a star to the unknown elements in a star. Hope this help!
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Give a chemical equation for aerobic cellular respiration
Aleonysh [2.5K]
C6H12O + 6OC2 + 6H2O + energy
5 0
3 years ago
1. What happens to the amount of energy as you move up the energy pyramid.
zepelin [54]

Answer:

Look it up

Explanation:

If you don't know then look it up

8 0
4 years ago
Read 2 more answers
An object is found to have a mass of 54.3 g. Find the object’s density, if its volume is 47.18 cm3
Schach [20]

Answer:

d≈ 1.15 g/cm^3

Explanation:

The density of an object can be found by dividing the mass over the volume.

d= m/v

The mass of the object is 54.3 grams and the volume is 47.18 cubic centimeters.

m= 54.3 g

v= 47.18 cm^3

Substitute the values into the formula.

d= 54.3 g/ 47.18 cm

Divide.

d= 1.150911403136922 g/cm^3

Let’s round to the nearest hundredth. The 0 in the thousandth place tells us to leave the 5 in the hundredth place.

d ≈ 1.15 g/cm^3

The density is about 1.15 grams per cubic centimeter.

8 0
3 years ago
Suppose 20.23 g of glucose are dissolved in 95.75 g of water at 27.0 OC. Glucose is nonvolatile (has no vapor pressure) and has
leonid [27]

Answer:

Explanation:

From the information given :

we can understand the solute is glucose and the solvent is water,

So, the weight of glucose = 20.23 g

the molecular weight of glucose = 180.2 g/mol

weight of water = 95. 75 g

the molecular weight of water = 18.02 g/mol

pure vapor pressure of water P_A = 26.7 \ mmHg at 27°C

moles of glucose = weight of glucose/ molecular weight of glucose

= 20.23/180.2

= 0.11 mole

moles of water =  weight of water / molecular weight of water

= 95.75/18.02

= 5.31 mole

mole fraction of glucose X_{glucose} = (moles of glucose)/(moles of glucose+ moles of water)

X_{glucose} = 0.11/(0.11 + 5.31)

X_{glucose} =  0.0203

mole fraction of glucose X_{water} = (moles of water)/(moles of water+ moles of glucose)

X_{water} = 5.31/ (5.31 + 0.11)

X_{water} =  0.9797

Using Raoult's Law:

P_S = P^0_A \times X_A \  \ \  OR  \  \ \  P_A = P^0_A \times X_A

where:

P_S = vapor pressure of the solution

P_A = total vapor pressure of the solution

P^0_A= vapor pressure of the solvent in the pure state

X_A = mole fraction of solvent i.e. water

P_A = 95.75 ×  0.9797

P_A =  93.81 mmHg

the total vapor pressure of the solution = 93.81 mmHg

4 0
3 years ago
The motion of objects can be quite complex. The motion of this frog is an example. Suppose that the frog moves from one lily pad
olga2289 [7]

Answer:

The speed of the frog is 1m per second the velocity is he can turn and jump in a different direction in 1 second the distance is 1m and there is no displacement. I'm pretty sure this is right

Explanation:

8 0
3 years ago
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