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3 years ago

A. Phenol (C6H5OH) is an aromatic compound and a colourless liquid widely used in household products

1 answer:

8 0

Because there are 2 Cl on the left, we will put a coefficient 2in front of HCl on the right side to balance out the Cl. This would result in an unequal amount of H, with 6 on the right side and 7 in the left, so we have to put a coefficient of 2 in front of C6H5OH and C6H4OH on both sides to balance out the H. By doing this, we would obtain an equal amount of H on both sides. The Carbon is already balanced, and so is the Oxygen.

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Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H,

ludmilkaskok [199]

Answer: The empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

Explanation:

We are given:

Percentage of H = 5.80 %

Percentage of O = 23.02 %

Percentage of N = 20.16 %

Percentage of Cl = 51.02 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 5.80 g

Mass of O = 23.02 g

Mass of N = 20.16 g

Mass of Cl = 51.02 g

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

Step 1: Converting the given masses into moles.

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.80g}{1g/mole}=5.80moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{23.02g}{16g/mole}=1.44moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{20.16g}{14g/mole}=1.44moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{51.02g}{35.5g/mole}=1.44moles$

Step 2: 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 1.44 moles.

For Hydrogen = $\frac{5.80}{1.44}=4.03\approx 4$

For Oxygen = $\frac{1.44}{1.44}=1$

For Nitrogen = $\frac{1.44}{1.44}=1$

For Chlorine = $\frac{1.44}{1.44}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of H : O : N : Cl = 4 : 1 : 1 : 1

Hence, the empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

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3 years ago

How do you balance the equation in number one?

AleksAgata [21]

____NaNO3 + ___PbO --> ___Pb(NO3)2 + ___Na[2]O

To balace the eqaution, you need to have the same number of atoms for each element on both the reactant (left) and product (right) side.

To start off, you wanna know the number of atoms in each element on both sides, so take it apart:

[reactants] [product]
Na- 1 Na- 2

N- 1 N- 2(it's 2 because the the subscript [2] is outside of the parenthesis)

O- 4 O- 7 (same reason as above)

Pb- 1 Pb- 1

Na is not balanced out, so add a coefficient to make it the same on both sides.In this case, multiply by 2:

2NaNO3

Now Na is balanced, but the N and O are also effected by this, so they also have to be multiplied by 2 and they become:

Na- 2 Na- 2

N- 2 N- 2 (it balanced out)

O- 7 (coefficient times subscript, plus lone O) O- 7 (balanced out)

Pb was already balanced so no need to mess with it, just put a 1 where needed (it doesn't change anything).

Now to put it back together, it will look like this:

2NaNO3 + 1PbO --> 1Pb(NO3)2 + 1Na[2]O

3 0

4 years ago

Q: A 25.5 mL aliquot of HCl (aq) of unknown concentration was titrated with 0.113 M NaOH (aq). It took 51.2 mL of the base to re

blondinia [14]

M ( HCl ) = ?

V ( HCl ) = 25.5 mL in liters : 25.5 / 1000 => 0.0255 L

M ( NaOH ) = 0.113 M

V ( NaOH ) = 51.2 mL / 1000 => 0.0512 L

number of moles NaOH:

n = M x V

n = 0.113 x 0.0512 => 0.0057856 moles of NaOH

mole ratio:

HCl + NaOH = NaCl + H2O

1 mole HCl -------------- 1 mole NaOH
( moles HCl ) ----------- 0.0057856 moles NaOH

(moles HCl ) = 0.0057856 x 1 / 1

= 0.0057856 moles of HCl

M ( HCl ) = n / V

M = 0.0057856 / 0.0255

= 0.227 M

Answer A

hope this helps!

4 0

3 years ago

Read 2 more answers

Calculate the number of sucrose molecules in a 75.0 gram sample.

zzz [600]

Answer:

1.32*10^23 molecules

Explanation:

sucrose formula: C12H22O11

molar mass: 12(12.01)+22(1.01)+11(16.00)=342.34g/mol

75.0 g C12H22O11 * (1 mol C12H22O11)/(342.34g C12H22O11)=0.219 mol C12H22O11

0.219 mol * (6.022*10^23)/mol = 1.32*10^23 molecules (three sig. figures)

5 0

3 years ago

The temperature of a 95.4 g piece of Cu increases from 25.0 °C to 48.0 °C when the Cu absorbs 849 J of heat. What is the specifc

melisa1 [442]

<h3>Answer:</h3>

0.387 J/g°C

<h3>Explanation:</h3>

To calculate the amount of heat absorbed or released by a substance we need to know its mass, change in temperature and its specific heat capacity.
Then to get quantity of heat absorbed or lost we multiply mass by specific heat capacity and change in temperature.
That is, Q = mcΔT

in our question we are given;

Mass of copper, m as 95.4 g

Initial temperature = 25 °C

Final temperature = 48 °C

Thus, change in temperature, ΔT = 23°C

Quantity of heat absorbed, Q as 849 J

We are required to calculate the specific heat capacity of copper

Rearranging the formula we get

c = Q ÷ mΔT

Therefore,

Specific heat capacity, c = 849 J ÷ (95.4 g × 23°C)

= 0.3869 J/g°C

= 0.387 J/g°C

Therefore, the specific heat capacity of copper is 0.387 J/g°C

3 0

3 years ago