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dolphi86 [110]
3 years ago
14

What is a goal of the planned mission MIRI?

Chemistry
2 answers:
astraxan [27]3 years ago
6 0

Answer:

The answer is B. to explore the Kupier Belt

Deffense [45]3 years ago
6 0

Answer:

The correct answer is option B, that is, to explore the Kupier belt.

Explanation:

The MIRI or the Mid-Infrared Instrument possesses both a spectrograph and a camera, which observes light in the mid-infrared region of the electromagnetic spectrum, exhibiting the longer wavelengths that are generally not seen with the naked eyes.  

The MIRI encompasses the range of wavelength in between 5 to 28 microns. The sensitive detectors present in them permits to observe the newly developing stars, the redshifted light of far-off galaxies, and faintly visible comets and the objects present in the Kupier belt.  

You might be interested in
A) What is the maximum number of grams of nickel bromide that can be produced from the reaction of 67.8 g of nickel with 37.3 g
svetoff [14.1K]

Answer:

The answer to your question is a) 51.07 g of NiBr₂   b) Nickel, 54 g

Explanation:

Data

mass of NiBr₂ = ?

mass if Ni = 67.8 g

mass of Br = 37.3 g

Balanced chemical reaction

                Ni  +  Br₂   ⇒   NiBr₂

Process

1.- Find the atomic mass of the reactants and the molar mass of the product

Ni = 59 g

Br = 79.9 x 2 = 159.8 g

NiBr₂ = 59 + 159.8 = 218.8 g

2.- Find the limiting reactant

theoretical yield  Ni/Br₂ = 59/159.8 = 0.369

experimental yield Ni/Br₂ = 67.8/37.3 = 1.81

The limiting reactant is Bromine because the experimental yield was lower than the theoretical yield.

3.- Calculate the mass of NiBr₂

                    159.8 g of Br₂ --------------- 218.8 g of NiBr₂

                      37.3 g of Br₂ --------------  x

                          x = (37.3 x 218.8) / 159.8

                          x = 8161.24/159.8

                          x = 51.07 g of NiBr₂

4.- Find the excess reactant

The excess reactant is Nickel

                59 g of Ni ---------------- 159.8 g of Br₂

                  x               ----------------  37.3 g of Br₂

                            x = (37.3 x 59)/159.8

                            x = 2200.7/159.8

                            x = 13.77 g of Ni

Excess Ni = 67.8 - 13.77

                 = 54 g

5 0
3 years ago
I need help ASAP! How many reactants and products are in each element according to the chemical equation above. Please explain t
Nikitich [7]
Answer: 1,4,1 for the reactants
3,1,4,1 for the products

Explanation:
The products are equal to the reactants because of the conservation of matter.
6 0
2 years ago
The average rate of consumption of br− is 1.86×10−4 m/s over the first two minutes. what is the average rate of formation of br2
Scorpion4ik [409]
By considering the reaction equation is:
5Br(aq)+BrO3(aq)+6H(aq)= 3Br2(aq)+3H2O(l)
when the average rate of consumption of Br = 1.86x10^-4 m/s
So from the reaction equation 
5Br → 3Br2 when we measure the average rate of formation (X) during the same interval So,
∴ 1.86x10^-4/5 = X / 3
∴X = 1.1 x 10^-4 m/s
∴the average rate of formation of Br2 = 1.1x10^-4 m/s


7 0
2 years ago
Who believed that a grain of sand could be divided indefinitely?
Ilya [14]

Answer:

Democritus

Explanation:

4 0
3 years ago
During the combustion of 2.00 g of coal, the temperature of 500 g of water inside the calorimeter increased from 25.0°c to 43.7°
pantera1 [17]
Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.
6 0
3 years ago
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