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tatyana61 [14]
3 years ago
6

Which is a chemical property of soda ash

Chemistry
2 answers:
Jobisdone [24]3 years ago
8 0

Answer:

The answer would be A. It's a highly basic substance.

Katen [24]3 years ago
5 0
It's a highly basic substance. 


Good luck! (:
You might be interested in
Consider the following data on some weak acids and weak bases
SpyIntel [72]

Answer:

b < c < a < d

Explanation:

The weak acid with the lowest pKa will be the most acidic. In the other way, the conjugate base which the acid is weak will be strong.

The weak base with the lowest pKb will be the most basic. And the conjugate base of the weak base will be a strong acid.

Ka Acetic acid = 1.8x10-5

Ka HCN = 1.9x10-10

Kb pyridine = 1.7x10-9

Kb NH3 = 1.8x10-5

NH4Br is the conjugate base of a weak base. That means is a strong acid.

NH4Br has the lowest pH

NaBr is the conjugate base of a strong acid, HBr. That means NaBr is neutral

The most basic between the conjugate base of the acetic acid, NaCH3CO2 and KCN is KCN because the acetic acid is the stronger acid regard to HCN.

The rank is:

NH4Br < NaBr < NaCH3CO2 < KCN

<h3>b < c < a < d</h3>
6 0
3 years ago
6) (a) Calculate the absorbance of the solution if its concentration is 0.0278 M and its molar extinction coefficient is 35.9 L/
Anvisha [2.4K]

Answer:

6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

6) (a) Absorbance

Beer's Law is

A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}

(b) Percent transmission

A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}

7) Absorbance

A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}

8 0
3 years ago
The coolant in automobiles is often a 50/50 % by volume mixture of ethylene glycol, C2H6O2, and water. At 20°C, the density of e
Misha Larkins [42]

Explanation:

Let the volume of the solution be 100 ml.

As the volume of glycol = 50 = volume of water

Hence, the number of moles of glycol = \frac{mass}{molar mass}

                                                  = \frac{density \times volume}{molar mass}

                         = \frac{1.1088 \times 50}{62 g/mol}

                         = 0.894 mol

Hence, number of moles of water = \frac{50 \times 0.998}{18}

                                             = 2.77

As glycol is dissolved in water.

So, the molality = 0.894 \times \frac{1000}{49.92}

                           = 17.9

Therefore, the expected freezing point = -1.86 \times 17.9

                                                                  = -33.31^{o}C

Thus, we can conclude that the expected freezing point is -33.31^{o}C.

6 0
3 years ago
The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

5 0
3 years ago
What is the interval used on the y-axis of this graph?
bezimeni [28]

Answer:

5

Explanation:

the numbers go up by 5

7 0
3 years ago
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