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natta225 [31]
3 years ago
15

Which item is dull.

Chemistry
2 answers:
DaniilM [7]3 years ago
7 0
A cloth sock is dull; it doesn’t shine
asambeis [7]3 years ago
7 0
A cloth sock because it doesn’t shine
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Write the balanced NET IONIC equation for the reaction that occurs when perchloric acid and calcium hypochlorite are combined. I
Elanso [62]

Answer:

H^+(aq)+(ClO)^-(aq)\rightarrow HClO{(aq)}

Explanation:

Hello there!

In this case, since perchloric acid is HClO4 and is a strong acid and calcium hypochlorite is Ca(ClO)2, the undergoing molecular chemical reaction turns out:

2HClO_4{(aq)}+Ca(ClO)_2{(aq)}\rightarrow 2HClO{(aq)}+Ca(ClO_4)_2{(aq)}

Thus, since the resulting hypochlorous acid is weak, it does not fully ionize, so it remains unionized, however, we can write the ions for the other species:

2H^++2(ClO_4)^-+Ca^{2+}+2(ClO)^-\rightarrow 2HClO{(aq)}+Ca^{2+}+2(ClO_4)^-

Now, we can cancel out the spectator ions, calcium and perchlorate, to obtain:

2H^+(aq)+2(ClO)^-(aq)\rightarrow 2HClO{(aq)}\\\\H^+(aq)+(ClO)^-(aq)\rightarrow HClO{(aq)}

Best regards!

3 0
3 years ago
Which of the following best explains why solids do not change
Lesechka [4]

Answer: The particles in a solid are packed very close to each other.

8 0
3 years ago
A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2
a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=13.00g

Mass of H_2O=2.662g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, \frac{12}{44}\times 13.00=3.54g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, \frac{2}{18}\times 2.662=0.296g of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = \frac{0.295}{0.295}=1

For Hydrogen = \frac{0.296}{0.295}=1

For Oxygen = \frac{0.603}{0.295}=2.044\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is CHO_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Hence, the empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

3 0
3 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
Read 2 more answers
What did you include in your response?<br>check all that apply ​
erma4kov [3.2K]

Answer: i don't know

Explanation:

u gave no information on what you're asking

6 0
3 years ago
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