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3 years ago

What is the pH of a solution that has a [OH-] 2.65*10 -9 M

Chemistry

1 answer:

PSYCHO15rus [73]3 years ago

7 0

Answer:

Perchloric acid is one of the six common strong acids, which means 100% dissociation in water. Thus, [HClO4] = [H+].

pH = -log 2.65 x 10^-5 = 4.58

Explanation:

I am new, and I am giving this a try. Hope this helps : )

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Tartaric acid is the white, powdery substance that coats sour candies such as sour patch kids. combustion analysis of a 12.01-g

shusha [124]

First we calculate the mass of carbon, hydrogen and oxygen.

Mass of Carbon;
Molar mass of CO₂ is 44.009g/mol and molar mass of C = 12.011 g/mol

Mass of Carbon in 14.08g CO2₂ = 12.011/44.009 x 14.08 = 3.84

Mass Hydrogen;
Molar mass H₂O is 18.0015g/mol and molar mass of H is 1.008and H₂ = 2.016

Mass Hydrogen in 4.32g of H₂O = 2.016/18.0015 x 4.32 = 0.48

Mass of Oxygen = total mas of tartaric acid – (mass of carbon + mass of hydrogen) = 12.01 - (3.84+0.48) = 7.69
Now divide carbon, hydrogen and oxygen through their respective atomic masses: 
C = 3.84/12.011 = 0.32
H = 0.48/1.008 = 0.48
O = 7.69/15.999 = 0.48
Now divide by smallest: 
C = 0.32/0.32 = 1 
H = 0.48/0.32 = 1.5
O = 0.48/0.32 = 1.5
multiply by 2 to get whole integers: 
C = 1 x 2 = 2
H= 1.5 x 2 = 3
O= 1.5 x 2 = 3
Thus, the Empirical formula of tartaric acid = C₂H₃O₃

4 0

3 years ago

A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting i

Kisachek [45]

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

$\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}$

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

$\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}\\$

$\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J$\cdot$K$^{-1}$g$^{-1}$}\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J$\cdot$ K$^{-1}$g$^{-1}$}\times \Delta T_{\text{w}}\\\text{47.7 J$\cdot$K$^{-1}$}\times \Delta T_{\text{Cu}}& = &-\text{585 J$\cdot$ K$^{-1}$g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}$

The final temperature of the water would be 35.5 °C.

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3 years ago

1. How many covalent bonds are expected to form from phosphorus in order to obey the octet rule?

Grace [21]

To satisfy the octet rule, there has to be eight electrons in phosphorus's outermost shell. Phosphorus has five electrons in its valence shell, fluorine has seven. This means that phosphorus would have to bond to three fluorine atoms in order to have eight electrons in its valence shell without breaking the octet rule in any one of the fluorine atoms.

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4 years ago

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umka21 [38]

The answer is natural gasses

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