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DaniilM [7]
3 years ago
6

Figure 1 shows a seaside cliff. Figure 2 shows the same cliff after a period of time has passed. What caused the change from fig

ure 1 to figure 2?
Chemistry
2 answers:
NeX [460]3 years ago
7 0
The change from figure one to figure two was most likely caused by erosion. Erosion is the process of something being eroded by wind, water, or other natural agents.
Romashka [77]3 years ago
5 0

The answer is rainfall and ocean waves. Hope I Helped!

You might be interested in
ASAP , 8.81 g Carbon
xz_007 [3.2K]

The empirical formula : C₂Cl₇

The molecular formula : C₁₀Cl₃₅

<h3>Further explanation</h3>

Given

8.81 g Carbon

91.2 g Chlorine

Molar Mass: 1362.5 g/mol

Required

The empirical formula and molecular formula

Solution

Mol ratio :

C = 8.81 g : 12.011 g/mol =0.733

Cl = 91.2 g : 35,453 g/mol = 2..572

Divide by 0.733

C : Cl = 1 : 3.5 = 2 : 7

The empirical formula : C₂Cl₇

(The empirical formula)n = the molecular formula

(C₂Cl₇)n = 1362.5

(2x12.011+7x35.453)n=1362.5

(272.193)n=1362.5

n = 5

6 0
3 years ago
1. Calculate the heat change associated with cooling a 350.0 g aluminum bar from
Amanda [17]

Answer:

14175 j heat released.

Explanation:

Given data:

Mass of aluminium = 350.0 g

Initial temperature = 70.0°C

Final temperature = 25.0°C

Specific heat capacity of Aluminium = 0.9 j/g.°C

Heat changed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Heat change:

ΔT = Final temperature - initial temperature

ΔT = 25.0°C - 70°C

ΔT = -45°C

Q = m.c. ΔT

Q = 350 g × 0.9 j/g.°C  × -45°C

Q = -14175 j

7 0
3 years ago
Why it’s possible to walk through air but not a concrete wall
Sunny_sXe [5.5K]

well you can't walk through the air but you can't walk through a concrete wall because all of the solid atoms holding it together

3 0
3 years ago
Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2
Marizza181 [45]

Answer:

\Delta _rH=-1124.14kJ/mol

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}

In such a way, by using the NIST database, we find that:

\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol

Thus, we plug in the enthalpies of formation to obtain:

\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol

Best regards!

8 0
3 years ago
The time necessary for the decay of one- half sample of a radioactive substance is a _____. mass defect daughter half life nucli
Mazyrski [523]

Answer: half life

Explanation: Radioactive decay follows first order kinetics and the time required for the decay of a radioactive material is calculated as follows:

t=\frac{2.303}{k}\hspace{1mm}log\frac{x}{a}

t= time required

k= disintegration constant

x= amount of substance left after time t

a= initial amount of substance

when one half of the sample is decayed, one half of the sample remains and t can be represented as t_{1/2}

at t= t_{1/2}, x=\frac{a}{2}

t_{1/2}=\frac{2.303}{k}\hspace{1mm}log\frac{a/2}{a}

t_{1/2}=\frac{0.693}{k}

3 0
3 years ago
Read 2 more answers
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