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aliya0001
1 month ago
6

How can you tell if a sample contains baking soda and cornstarch or baking powder?

Chemistry
2 answers:
VMariaS [2.9K]1 month ago
7 0
You can test it by combining it with vinegar. The acetic acid present in vinegar reacts with sodium bicarbonate (baking soda) to produce carbon dioxide, resulting in an intense bubbling reaction that is harmless. Baking powder, however, won’t produce this effect.
KiRa [2.9K]1 month ago
4 0

Explanation:

Hi there! Let’s figure this out!

To determine if the sample contains baking soda, we can introduce vinegar.

Mixing vinegar (acetic acid) with sodium bicarbonate leads to the creation of water, carbon dioxide, and sodium acetate, resulting in visible bubbles (the carbon dioxide gas).

You might be interested in
An air sample consists of oxygen and nitrogen gas as major components. It also contains carbon dioxide and traces of some rare g
Anarel [2989]

Explanation:

A mixture refers to a substance that contains two or more different kinds of substances that are combined physically.

For instance, air is a mixture that contains oxygen, nitrogen, and other gases.

A heterogeneous mixture is characterized by uneven distribution of solute particles in the solvent.

For example, sand suspended in water is a heterogeneous mixture.

Conversely, a homogeneous mixture is one where the solute particles are uniformly distributed in a solvent.

A homogeneous mixture appears as a clear solution.

For instance, when salt dissolves in water, it forms a homogeneous mixture.

A solution is defined as a mixture of two or more substances combined together.

A compound consists of two or more different elements chemically bonded together in a specific mass ratio.

An element is the simplest form of a substance that comprises only one type of atom.

For example, a piece of sodium is composed solely of sodium atoms.

Conversely, a pure substance refers to a material consisting of just one type of molecule or atom.

For example, O_{2}, N_{2} etc are considered pure substances.

Thus, it can be concluded that the air sample can be described using the terms:

  • pure chemical substance.
  • heterogeneous mixture.
  • mixture.
4 0
26 days ago
What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c
Tems11 [2777]

Solution:

The molecular formula is PbSO₄, indicating lead sulfate

Option c.

Explanation:

The percentage makeup shows that in 100 g of this compound, there are:

68.3 g of Pb, 10.6 g of S, and (100 - 68.3 - 10.6) = 21.1 g of O

To find the moles of each element, we divide by their molar masses:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

Next, we find the mole ratio by dividing each by the smallest number of moles:

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Thus, the molecular formula is PbSO₄, representing lead sulfate.

8 0
1 month ago
Read 2 more answers
An experimental drug, D, is known to decompose in the blood stream. Tripling the concentration of the drug increases the decompo
lions [2927]

Answer:

The rate law for the decomposition reaction is:

R=k[D]^2

The unit for the rate constant will be M^{-1}s^{-1}

Explanation:

D\rightarrow Product

The rate law can be expressed as:

R=k[D]^x..[1]

When the drug concentration is tripled, the decomposition rate rises by a factor of nine.

[D]'=3[D]

R'=9\times R

R'=k[D]'^x...[2]

[1] ÷ [2]

\frac{R}{R'}=\frac{k[D]^x}{k[D']^x}

\frac{R}{9R}=\frac{k[D]^x}{k[3D]^x}

9=3^x

Solving for x results in:

x = 2.

This indicates a second-order reaction.

The decomposition reaction's rate law is:

R=k[D]^2

The unit for the rate constant will be:

k=\frac{R}{[D]^2}=\frac{M/s}{(M)^2}=M^{-1}s^{-1}

The unit for the rate constant will be M^{-1}s^{-1}.

5 0
1 month ago
En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello
KiRa [2933]

Answer:

a) m_{CH_4}=2630kg

b) 1657 €

Explanation:

Hola,

a) En esta cuestión analizaremos el millón de litros de agua anualmente, dado que este dato nos permite calcular el calor necesario para calentar dicha cantidad, considerando que la densidad del agua es de 1 kg/L:

Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ

A continuación, utilizamos la entalpía de combustión del metano para determinar la cantidad en kilogramos necesaria, ya que la energía calórica perdida por el metano es equivalente a la energía obtenida por el agua:

Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}

m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg

b) En este supuesto, tenemos que, bajo condiciones normales de 1 bar y 273 K, el precio de 1 metro cúbico de metano es 0,45 €, lo que nos permite calcular las moles de metano en esas condiciones:

n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol

En consecuencia, los kilogramos de metano que se obtienen por 0,45 € son:

44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4

Finalmente, usando regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0
1 month ago
A characteristic feature of any form of chromatography is the ________.a. calculation of an Rf value for the molecules separated
castortr0y [3046]

Answer: The right choice is (c) application of both a mobile phase and a stationary phase.

Explanation:

Chromatography: This refers to a technique for separating a mixture where the mixture is distributed between two phases at varying rates, one being stationary and the other moving.

Mobile phase: The component in which the mixture is dissolved is referred to as the mobile phase.

Stationary phase: This is an adsorbent medium that remains in place while a liquid or gas passes over its surface, thus remaining stationary.

Consequently, a key characteristic of any chromatography technique involves utilizing both a mobile and a stationary phase.

4 0
1 month ago
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