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Degger [83]
2 years ago
14

Which phrase describes faults?

Chemistry
2 answers:
AleksAgata [21]2 years ago
6 0
Faults is cause gaps in the geologic record
Softa [21]2 years ago
4 0
Faults are gaps in the geologic record, but can also be called cracks that form in the Earths crust
You might be interested in
If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction
Tasya [4]

Answer:

<u></u>

  • <u>0.456M</u>

Explanation:

<u>1. Balanced molecular equation</u>

     2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O

<u>2. Mole ratio</u>

     \dfrac{2molHNO_3}{1molBa(OH)_2}

<u>3. Moles of HNO₃</u>

  • Number of moles = Molarity × Volume in liters
  • n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

<u>4. Moles Ba(OH)₂</u>

  • n = 0.700M × 0.0310 liter = 0.0217 mol

<u>5. Limiting reactant</u>

Actual ratio:

   \dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

<u>6. Final molarity of Ba(OH)₂</u>

  • Molarity = number of moles / volume in liters
  • Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
5 0
3 years ago
What is the heat in joules required to melt 25 grams of ice?
AfilCa [17]

Answer:

8,350 Joules

Hope this helps

7 0
2 years ago
Two scientists wrote a paper detailing their research and conclusions and submitted it to a scientific journal. Several months l
Semmy [17]
<span>Two scientists wrote a paper detailing their research and conclusions and submitted it to a scientific journal. Several months later, they received the paper back from the publisher with many comments attached from several fellow scientists. It is either that they revise their study or replicate the study. Most scientists would revisit their work and the findings they had from their research. Most probable if they were successful and the comments of the publisher and the co-scientists were positive they could replicate the study to validate its accountability.<span>
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6 0
3 years ago
Write the balanced equation for the neutralization reaction between HCl and Ba(OH)2 in an aqueous solution. Phases are optional.
ad-work [718]

Explanation:

2HCl + Ba(OH)2 ------> BaCl2 + 2H2O

4 0
2 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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