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3 years ago

In this photograph what rock formation is presented by the blue lines?

Chemistry

2 answers:

svp [43]3 years ago

8 0

k12 right well its im in it as well earth science is hard anyways its syncline

also could you please mark brainliest

Effectus [21]3 years ago

3 0

It is representing syncline rock formation

there are two rock formation based on fold formation : syncline and anticline

In syncline rock formation there is fold like trough unlike anticline where it is like crust

In syncline the fold is downward as shown in photo and the new rock is outer fold and old at inner side

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Cindy made tea. She started with 300 grams of water at 20 degrees Celsius. She transferred 18,000 calories to the water. What wa

Serjik [45]

Answer:

T final = 80°C

Explanation:

Q = mCpΔT

∴ Q = 18000 cal

∴ m H2O = 300 g

∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K

∴ T1 = 20°C = 293 K

∴ T2 = ?

⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)

⇒ (18000 cal)/(300 cal/K) = T2 - 293 K

⇒ T2 = 293 K + 60 K

⇒ T2 = 353 K (80°C)

8 0

4 years ago

If 11.00 mL of a standard 0.2831 M NaOH solution is required to neutralize 26.86 mL of H2SO4, what is the molarity of the acid s

MrRa [10]

Answer:

0.1159 M

Explanation:

Using the formula below:

CaVa = CbVb

Where;

Ca = concentration/molarity of acid (M)

Cb = concentration/molarity of base (M)

Va = Volume of acid (mL)

Vb = volume of base (mL)

According to this question, the following information were given:

Ca = ?

Cb = 0.2831 M

Va = 26.86 mL

Vb = 11.00 mL

Using CaVa = CbVb

Ca × 26.86 = 0.2831 × 11

26.86Ca = 3.1141

Ca = 3.1141 ÷ 26.86

Ca = 0.1159

The molarity of the acid (H2SO4) solution is 0.1159 M

3 0

3 years ago

Determine el porcentaje masa volumen de una solucion compuesta de 150g de etanol y 20g de agua cuyo volumen total es de 165ml

drek231 [11]

Answer:

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4 0

3 years ago

SICL4 + H2O sio2 hcl

atroni [7]

Do you want this equation balanced?

If so, it would be

SiCl4 + 2H2O ---> SiO2 + 4HCl

5 0

3 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago