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Nutka1998 [239]
2 years ago
7

How does Dr. Hayes' and Dr. Malaska’s research differ? Why are both research projects important?

Chemistry
1 answer:
kati45 [8]2 years ago
6 0

Answer:

Answer:  What can experiments in a lab tell us about substances on Titan? Experiments in a lab can tell us that the lake did not evaporate in 2007 because the molecular attraction was a lot stronger, then it got weaker overtime.

How does Dr. Hayes' and Dr. Malaska’s research differ? Why are both research projects important? Their research differs because they were both talking about different things, Hayes was talking about how many lakes there were, while Malaska's was doing more hands on stuff like experiments. Both are important because we need to learn how the lakes formed, but we also need to do hands on experiments.

Explanation:

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Heat and pressure deep beneath earths surface can change any rock into what?
Alik [6]

Answer:

Metamorphic

Explanation:

Because it describes as a rock framed from other rocks by the action of heat and pressure.

5 0
3 years ago
How many grams of Br are in 335g of CaBr2
ASHA 777 [7]
The answer is 267.93 g

Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol

The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%

Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
7 0
3 years ago
During chemical reactions the bonds between atoms break and new bonds form. Energy must be absorbed to break a bond, so breaking
ss7ja [257]

Answer:

The equation is: CuCO3(s) → CuO(s) + CO2(g)

Explanation:

Copper carbonate decomposes at high temperatures, generating the products carbon dioxide (CO2) and copper oxide (II) CuO. In this type of decomposition reaction, a substance is broken generating two different compounds.

7 0
3 years ago
Calculate δg o for each reaction using δg of values:(a) h2(g) + i2(s) → 2hi(g) kj (b) mno2(s) + 2co(g) → mn(s) + 2co2(g) kj (c)
steposvetlana [31]
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol

Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol

Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
      = (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
      = 176.15 kJ - 84.78 kJ = 91.38 kJ 
 





 
6 0
3 years ago
Read 2 more answers
Give two examples of diffusion process in plants and animals.
lisabon 2012 [21]

answer:

in plants

Transport manufactured food from the leaves to others parts of the plant

Facilitates gaseous exchange through the stomata in the leaves to other parts of the plant

in animal

Exchange of respiratory gases across respiratory services

Excretion of nitrogenous waste in some unicellular organisms

Explanation:

Hope it benefit

3 0
3 years ago
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