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3 years ago

In which type of condition is a conditions is a landslide most likely to occur

Chemistry

1 answer:

masha68 [24]3 years ago

6 0

Answer:

B. In a wet, hilly area

Explanation:

We know that for a landslide to happen, there has to be a down hill slope so it can fall.

This eliminates answers C, D and E.

So we are left with A and B

Now we can look at this in a real life scenario. If you have ever walking on wet ground, you will know that is weaker than dry ground. Knowing this, we know the answer will have to be answer B.

You might be interested in

What is the pH of an aqueous solution with a hydrogen ion concentration of [H ] = 8.2 × 10–4 M?

stira [4]

PH = - log [ H+ ]

pH = - log [ 8.2 x 10⁻⁴]

pH = 3.08

hope this helps!

7 0

3 years ago

What is the Kc for the following reaction at equilibrium at 500∘C if [N2]= 0.41 M , [H2]= 0.41 M , and [NH3]= 2.2 M ? N2(g)+3H2(

viva [34]

Answer:

answer po Yan hope it help

6 0

3 years ago

A student titrated a solution containing 3.7066 g of an unknown Diprotic acid to the end point using 28.94 ml of 0.3021 M KOH so

uranmaximum [27]

Answer:

Molar mass of the diprotic acid is 424 grams

Explanation:

[hint: diprotic acid only contains 2 hydrogen protons]

Ionic equation:

${ \bf{2OH { }^{ - } _{(aq)} + 2H { }^{ + } _{(aq)}→ 2H _{2} O _{(l)} }}$

first, we get moles of potassium hydroxide in 28.94 ml :

${ \sf{1 \: l \: of \: KOH \: contains \: 0.3021 \: moles}} \\ { \sf{0.02894 \: l \: of \: KOH \: contain \: (0.02894 \times 0.3021) \: moles}} \\ { \underline{ = 0.008743 \: moles}}$

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.

Therefore, moles of acid that reacted are 0.008743 moles.

${ \sf{0.008743 \: moles \: of \: acid \: weigh \: 3.7066 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 3.7066}{0.008743}) \: g }} \\ = { \underline{423.95 \: g \approx424 \: grams}}$

for the molar mass:

8 0

3 years ago

Enter your answer in the provided box. The vapor pressure of ethanol is 1.00 x 10² mmHg at 34.90°C. What is its vapor pressure a

zavuch27 [327]

Answer:

2,54x10² mmHg

Explanation:

To solve this problem you can use Clausius-Clapeyron equation that serves to estimate vapor pressures or temperatures:

$Ln(\frac{P_{2}}{P_{1}} ) =\frac{ deltaH_{vap}}{R} (\frac{1}{T_{1}}-\frac{1}{T_{2}} )$

Where:

P1 is 1,00x10² mmHg

ΔHvap is 39,3 kJ/mol

R is gas constant 8,314x10⁻³ kJmol⁻¹K⁻¹

T1 is 34,90°C + 273,15 = 308,05 K

T2 is 54,81°C + 273,15 = 327,96 K

Thus:

$Ln(\frac{P_{2}}{1,0x10^{2}mmHg}) =\frac{39,3kJ/mol}{8,314x10^{-3}kJ/molK} (\frac{1}{308,05K}-\frac{1}{327,96K} )$

Thus, P2 is 2,54x10² mmHg

I hope it helps!

3 0

4 years ago

Please help with #2 and #3

Vaselesa [24]

Answer:

2. $V_2=17L$

3. $V=82.9L$

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

$\frac{V_2}{T_2}=\frac{V_1}{T_1}$

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

$V_2=\frac{V_1T_2}{T_1}$

Therefore, we plug in the given data to obtain:

$V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L$

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

$V=\frac{nRT}{P}$

Therefore, we plug in the data to obtain:

$V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L$

Best regards!

6 0

3 years ago