which feature of the ocean floor includes its deeper parts ? A.Trench B.estuary C.mid-ocean D. contnetal slop

What is the average range of the PH scale?<br><br> THANKS!!

alex41 [277]

<span>The pH scale goes from 0-14. 0-6.9 is acidic, 7 is neutral and 7.1-14 is basic</span>

8 0

3 years ago

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1. How many molecules of S2 gas are in 756.2 L?

AfilCa [17]

Answer: There are $2.032 \times 10^{25}$ molecules $S_{2}$ gas are in 756.2 L.

Explanation:

It is known that 1 mole of any gas equals 22.4 L at STP. Hence, number of moles present in 756.2 L are calculated as follows.

$Mole = \frac{Volume}{22.4 L}\\= \frac{756.2 L}{22.4 L}\\= 33.76 mol$

According to mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ molecules.

Therefore, molecules of S present in 33.76 moles are calculated as follows.

$1 mol = 6.022 \times 10^{23}\\33.76 mol = 33.76 \times 6.022 \times 10^{23}\\= 2.032 \times 10^{25}$

Thus, we can conclude that there are $2.032 \times 10^{25}$ molecules $S_{2}$ gas are in 756.2 L.

5 0

3 years ago

10 pointss..******^^

Kisachek [45]

Answer:

close off completely

Explanation:

get wider

7 0

2 years ago

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A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

3 0

2 years ago

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Calculate the mole fraction of cai2 in an aqueous solution prepared by dissolving 0.400 moles of cai2 in 850.0 g of water.

alukav5142 [94]

1) Formulas:

a) mole fraction of component 1, X1

X1 = number of moles of compoent 1 / total number of moles

b) Molar mass = number grams / number of moles => number of moles = number of grams / molar mass

2) Application

Number of moles of CaI2 = 0.400

Molar mass of water = 18.0 g/mol

Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol

Total number of moles = 0.400 + 47.22 =47.62

Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840

7 0

2 years ago