Kichi made a table to identify where she might find acids and bases in her house based on their common uses. Substance ammonium hydroxide sulfuric acid citric acid potassium hydroxide Location cleaning cabinet cosmetics bag refrigerator flashlight What error did Kichi make?
Ammonium hydroxide would be found in the pantry because it is found in vinegar.

Sulfuric acid would be found in the garage because it is used in car batteries,

Citric acid would be found in the garage because it is used to clean rust from metal.
Potassium hydroxide would be found in the kitchen cabinet because it is used in oven cleaner.​

Answer: Alloy

Explanation: gold is an alloy, or mixture of metals

Answer:

[tex]c=0.2\ J/g^{\circ} C[/tex]

Explanation:

Given that,

Mass of a glass, m = 4 g

Initial temperature, [tex]T_i=5^{\circ} C[/tex]

Final temperature, [tex]T_f=45^{\circ} C[/tex]

Heat absorbed, Q = 32 J

We need to find the specific heat of the glass. The formula for the heat absorbed is given by :

[tex]Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{32\ J}{4\ g\times (45-5)^{\circ} C}\\\\=0.2\ J/g^{\circ} C[/tex]

So, the required specific heat of the glass is [tex]0.2\ J/g^{\circ} C[/tex].

The answer is A. The vibration caused by the waves through the air eventually weaken, which is why sound diminishes easily over distance.

The standard reduction potential for the given question is-1.68 V, under the condition that the given half-reaction is  Al(aq) +3e→ Al(s), then the required answer to the given question is Option B.

The standard reduction potential refers to the potential difference comparing the standard hydrogen electrode (SHE) and the half-reaction. The (SHE) possess a standard reduction potential of 0 V at all temperatures.

Hence, the standard reduction potential for the half-reaction Al(aq) + 3e→ Al(s) can be evaluated as follows

E°(Al₃⁺(aq) + 3e⁻ → Al(s)) - E°(H⁺(aq) + e⁻ → 1/2H₂(g))

= -1.66 V - 0 V

= -1.66 V

≈ -1.68 V

A half-reaction can occur for either the oxidation or reduction reaction component of a redox reaction. It projects the change in oxidation states of the substances involved. Half-reactions are applied to describe the processes in an electrochemical cell, in which one metal is oxidized at the anode and another metal is reduced at the cathode.

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Answer:

B

Explanation:

-1.68 V

Answer:

The Atomic Number

Explanation:

Answer:

Non polar covlant

Explanation:

Answer:

Theory- 1 to 7

Explanation:

Exercise- 8 to 25

ect

The right alternative is Option D (0.78 V).

According to the question,

By putting the values, we get

Cu+2 + 2e → Cu E = 0.52 V

Ag+1 + 1e Ag E° = 0.78 V

= 1.23 - 0.45

= 0.78.

negative

The standard cell potential is quite negative, so the reaction will not occur spontaneously as written.

Missing parts of the question.

Using the two cell reduction potentials shown for their corresponding reaction, calculate the cell potential for a voltaic cell made from these two systems: 0.78.

Cu+2 + 2e → Cu E = 0.52 V

Ag+1 + 1e Ag E° = 0.80 V

options:

A)

0.28 V

B)

–1.32 V

C)

1.32 V

D)

–0.28 V

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Answer:false

Explanation:

Identify the type of solid from the description. Select from the following choices: metallic solid, network solid, molecular solid, ionic solid, or amorphous solid. a. This solid is ductile and conducts electricity.b. This solid is not conductive and has a low melting point.

Answer:

See explanation

Explanation:

A is a Metallic Solid

Let us look again at the physical properties of metals;

Metals are

We can now see that both ductility and electrical conductivity are metallic properties.

B is a Molecular solid

Molecules are formed when atoms share electrons. In molecules, there are no free electrons hence molecules do not conduct electricity.

Secondly, molecules are held together in the solid state by weak dispersion forces hence molecular solids have a low melting point.

Answer:

D. Oxidation

Explanation:

In an electrolytic cell, oxidation (the loss of electrons) takes place at the anode. At the cathode reduction takes place.

A mnemonic technique that could be used is that the process that starts with a vowel (Oxidation) takes place at the place that also starts with a vowel (Anode).

Answer:

The relationship is the "variation of vapour pressure with temperature" or " the effect of vapour pressure on the melting and boiling points in a phase"

true  becouse i sed so                                            

Answer:

B) 2

Explanation:

There are two significant figures in given measurement,

2, 0

Significant figures:

1= The given measurement have four significant figures 1534.

2= All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.

3= Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.

4= Zero between the non zero digits are consider significant like 107 consist of three significant figures.

5= The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.

Answer:

They are all mixture

Explanation:

last option

Answer:

[tex]c=0.127\ J/g^{\circ} C[/tex]

Explanation:

Given that,

Mass of the sample, m = 275 g

It required 10.75 kJ of heat to change its temperature from 21.2 °C to its melting temperature, 327.5 °C.

We need to find the specific heat of the metal. The heat required by a metal sample is given by :

[tex]Q=mc\Delta T[/tex]

c is specific heat of the metal

[tex]c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{10.75\times 10^3\ J}{275\times (327.5 -21.2)}\\\\=0.127\ J/g^{\circ} C[/tex]

So, the specific heat of metal is [tex]0.127\ J/g^{\circ} C[/tex].

Answer:

The molar mass and molecular weight of C3H6O is 58.0791.

Explanation:

Answer:

[tex]PNCl_2[/tex]

Explanation:

Hello!

In this case, when determining empirical formulas by knowing the by-mass percent, we first must assume the percentages as masses so we can compute the moles of each element:

[tex]n_P=\frac{26.73g}{30.97g/mol}=0.863mol\\\\n_N=\frac{12.09g}{14.01g/mol}=0.863mol\\\\n_C_l=\frac{61.18g}{35.45g/mol}=1.726mol[/tex]

Now, for the determination of the subscript of each element in the empirical formula, we divide the moles by the fewest moles (P or N):

[tex]P=\frac{0.863mol}{0.863mol}=1\\\\N= \frac{0.863mol}{0.863mol}=1\\\\Cl=\frac{1.726mol}{0.863mol}2[/tex]

Thus, the empirical formula is:

[tex]PNCl_2[/tex]

Regards!

Answer:

Because the compound has a formula of M2O, the number of valence electrons of M should be 1. Therefore, (1) Group 1 is the correct answer because elements in group 1 have 1 valence electron.

Explanation:

Answer:

True

Explanation:

Answer: D an organism

Explanation:

The solubility of lead chromate : s= 5.29 x 10⁻⁷ mol/L

Given

Ksp  for PbCl,=2.8x10*-13 mole/dmº

PbCrO4

Required

The solubility

Solution

P b C r O ₄ ( s ) ⇌ P b²⁺( a q) +C r O ₄ ⁻²( a q )

s                     s               s

s= solubility

Ksp P b C r O ₄  = s²

K s p  = [ P b ²⁺] [C r O ₄ ⁻² ]= 2.8x10⁻¹³

s = √2.8x10⁻¹³

s= 5.29 x 10⁻⁷ mol/L

Answer:

2

Explanation:

because first shell ( k shell ) needs only 2 electron to complete its octet .

hope it helps