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maks197457 [2]
2 years ago
8

What causes the different colors of layers in a sedimentary rock? A. the cooling process of the magma B. the weather in the area

C. different minerals and sediments in the rock​
Chemistry
1 answer:
OLga [1]2 years ago
4 0

Different colors of sedimentary rock are determined by the environment where they are located. For example, red rocks form where oxygen is present. Darker sediments form when the environment is oxygen poor.

So, basically, the weather and surroundings of the rock, so B.

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How does a liquid substance change into gas? ​
Afina-wow [57]

Answer: heat

Explanation:

Is that heat boils the liquid and the liquid eventually evaporated into gas

3 0
3 years ago
A customer experiences worsening side effects in response to a prescription. What do you suggest to them?
o-na [289]

Answer:

Probably stop taking the prescribed durg and contact your pharmacist and your doctor that gave you your prescription asap.

Explanation:

Both of those health professionals will assist the patient in understanding how to go about the next steps for side effect relief.

8 0
2 years ago
]According to Schrodinger's work with electrons and their orbital paths, what can be determined about the electrons of an atom?
kumpel [21]
Schrodinger developed a famous equation that allows the solutions for electron wave functions to be found given a specific potential. For the case of an atom, Schroginger's equation allows the determination of electron wave functions. These wave functions tell us how electrons are distributed in space around the atom.
3 0
3 years ago
In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce ga
Reptile [31]

Answer:

The rate at which ammonia is being produced is 0.41 kg/sec.

Explanation:

N_2+3H_2\rightarrow 2NH_3 Haber reaction

Volume of dinitrogen consumed in a second = 505 L

Temperature at which reaction is carried out,T= 172°C = 445.15 K

Pressure at which reaction is carried out, P = 0.88 atm

Let the moles of dinitrogen be n.

Using an Ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.88 atm\times 505 L}{0.0821 atm l/mol K\times 445.15 K}=12.1597 mol

According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.

Then 12.1597 mol of dinitrogen will produce :

\frac{2}{1}\times 12.1597 mol=24.3194 mol of ammonia

Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol

=413.43 g=0.41343 kg ≈ 0.41 kg

505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.

6 0
3 years ago
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0
3 years ago
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