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Fantom [35]
3 years ago
13

Which of the following is the preferred degree for a seismologist?

Chemistry
1 answer:
Sedbober [7]3 years ago
5 0
<em>A seismologist is a scientist who does research in </em><span><em>seismology or history Which is same as <u>Geoscience</u>.</em></span><em></em>
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How many joules of heat are absorbed when 73 g water are heated from 30*C to 43*C? *
stepladder [879]

Answer:

3966.82 J

Explanation:

q=sm∆T

q=73×13×4.18

the specific heat for water is 4.18

6 0
3 years ago
Read 2 more answers
I need help ASAP........
Luba_88 [7]

Answer:

sodium hydroxide and hydrochloric acid is the reactants

7 0
3 years ago
Calculate the molarity of a solution made by dissolving 4.0 g of calcium bromide, CaBr2, in enough water to give 200 mL of solut
Troyanec [42]

Answer:

0.1 M

Explanation:

Molarity = number of moles / litres of solution.

4 g of calcium bromide = 0.02 mol

(found by dividing 4 g by the atomic mass of CaBr2, which is 199.886)

200 mL of solution = 0.2 litres

Molarity = 0.02 mol / 0.2 L = 0.1 M

8 0
3 years ago
Water is an essential molecule to life on Earth. Which of the following is not a property of water that is critical to biologica
Musya8 [376]
I believe it will be A. Density, temperature, and viscosity, can be found in water through tests. Hope this helps! Mark as brainly please!
7 0
2 years ago
Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
love history [14]

Answer:

(a) V_B=11.68L

(b) x_{He}=0.533

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol

Thus, since the final pressure is 3.60 bar, we can write:

P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar

The moles of helium could be computed via solver as:

n_{He}=2.358mol

Or algebraically:

3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol

In such a way, the volume of the compartment B is:

V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\  \\V_B=11.68L

Finally, he mole fraction of He is:

x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533

Regards.

8 0
3 years ago
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