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2 years ago

The bubbles that make a baked cake light and tender come from the decomposition of carbonic acid (H2CO3) in the cake batter.

Chemistry

2 answers:

igomit [66]2 years ago

5 0

Answer: $CO_2$ and $H_2O$

Explanation:

Decomposition reaction is a type of chemical reaction in which one reactant gives two or more than two products.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$H_2CO_3\rightarrow H_2O+CO_2$

ser-zykov [4K]2 years ago

3 0

I believe that carbonic acid breaks down into carbon dioxide and water when under decomposition.

The answer should be CO2 and H2O

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How much maleic anhydride would you need to react 178 mg of anthracene? Assume 1:1 ratio from maleic anhydride to anthracene.

Licemer1 [7]

Answer:

(1) 0.10 (2) 17.8 g

Explanation:

Since the reaction ratio is 1:1 what we need is to convert the given masses to moles and you will have the answer:

MW anthracene = 178.23 g/mol

MW maleic anhydride = 98.06 g/mol

a) mass anthracene = 178 mg x 1 g/ 1000 mg = 0.178 g anthracene

Moles anthracene = 0.178 g anthracene/ 178.23 g/mol

= 0.001 mol anthracene

0.001 mol anthracene x 1 mol maleic acid/mol anthracene

= 0.001 mol maleic anhydride

mass maleic anhydride = 0.001 mol x 98.06 g/mol = 0.10 g

b) moles maleic anhydride = 9.8 g/ 98.06 g/mol = 0.099 moles

0.099 moles maleic anhydride x 1 mol anthracene/mol maleic anhydride =

0.099 mol anthracene

g anthracene = 0.10mol x 178 g/mol = 17.8 g

8 0

2 years ago

What is the molarity of a naoh solution if 11.9 ml of a 0.220 m h2so4 solution is required to neutralize a 25.0-ml sample of the

bulgar [2K]

0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule.
So moles of protons = 0.01 x 2 = 0.02 moles of H+ 
For neutralization: moles H+ = moles OH- 
Therefore moles of NaOH = 0.02 
conc = moles / volume 
Conc NaOH = 0.02 / 0.025L = 0.8M

4 0

2 years ago

Read 2 more answers

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

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

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago

The interior of a refrigerator has a volume of 0.600 m3. the temperature inside the refrigerator in 282 k, and the pressure is 1

Dafna11 [192]

In this instance we can use the ideal gas law equation to find the number of moles of gas inside the refrigerator
PV = nRT
where
P - pressure - 101 000 Pa
V - volume - 0.600 m³
n - number of moles
R - universal gas constant - 8.314 J/mol.K
T - temperature - 282 K
substituting these values in the equation
101 000 Pa x 0.600 m³ = n x 8.314 J/mol.K x 282 K
n = 25.8 mol
there are 25.8 mol of the gas
to find the mass of gas
mass of gas = number of moles x molar mass of gas
mass = 25.8 mol x 29 g/mol = 748.2 g
mass of gas present is 748.2 g

6 0

2 years ago

The volume of a single strontium atom is 4.15×10-23 cm3. What is the volume of a strontium atom in microliters

Ivenika [448]

Answer:- $4.15*10^-^2^0\mu L$