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3 years ago

You have 1.0 mole of each compound below. which has the greatest mass?

1 answer:

kramer3 years ago

5 0

(a) Iron (iii) sulphate:
From the periodic table:
mass of iron = 55.845 grams
mass of sulphur = 32.065 grams
mass of oxygen = 16 grams
Iron (iii) sulphate has the formula: Fe2(SO4)3
molar mass = 2(55.845) + 3(32.065) + 3(4)(16) = 399.885 grams

(b) Sodium hydroxide:
From the periodic table:
mass of sodium = 22.989 grams
mass of oxygen = 16 grams
mass of hydrogen = 1 gram
Sodium hydroxide has the formula: NaOH
molar mass = 22.989 + 16 + 1 = 39.989 grams

(c) Barium carbonate
From the periodic table:
mass of barium = 137.327 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Barium carbonate has the formula: BaCO3
molar mass = 137.327 + 12 + 3(16) = 197.327 grams

(d) ammonium nitrate:
From the periodic table:
mass of nitrogen = 14 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
Ammonium nitrate has the formula: NH4NO3
molar mass = 14 + 4(1) + 14 + 3(16) = 80 grams

(e) Lead (iv) oxide
From the periodic table:
mass of lead = 207.2 grams
mass of oxygen = 16 grams
Lead (iv) oxide has the formula: PbO2
molar mass = 207.2 + 2(16) = 239.2 grams

From the above calculations, we can see that:
Iron (iii) sulphate has the greatest mass.

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a compound has 15.39 g of gold for every 2.77 g of chlorine. simplified there is _____ g of gold for every 1 g of chlorine

iren [92.7K]

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction $\frac{a}{b}$

You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:

$\frac{15.39 g of gold}{2.77 g of chlorine}$

The proportion is the equal relationship that exists between two reasons and is represented by: $\frac{a}{b}=\frac{c}{d}$

This reads a is a b as c is a d.

To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:

$\frac{15.39 g of gold}{2.77 g of chlorine}=\frac{mass of gold}{1 g of chlorine}$

Solving for the mass of gold gives:

$mass of gold=1 g of chlorine*\frac{15.39 g of gold}{2.77 g of chlorine}$

mass of gold= 5.56 grams

So, <u><em>there is 5.56 g of gold for every 1 g of chlorine</em></u>

5 0

3 years ago

What should you do if you spill a chemical on your hand?

dimulka [17.4K]

Answer:

You should rinse your head thoroughly ASAP, don't try to blow on the chemical or anything. Its going burn, you have to let your head cool off.

Explanation:

why rinse off your hand? you're getting the fluid away fast, so I can't infect your skin that much.

5 0

3 years ago

If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate

andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

Volume of NaOH = 32.40 mL = 0.03240 L

Molarity of NaOH = 0.258 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of the acetic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 1

⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.258 M

⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The concentration of acetic acid is 8.36 M

6 0

3 years ago

A solution contains 90 milliequivalents of HC1 in 450ml. What is its normality?

Rashid [163]

Answer:

Normality N = 0.2 N

Explanation:

Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.

Normality is represented by N.

Mathematically, we have :

$\mathbf{Normality \ N = \dfrac{Number \ of \ gram \of \ equivalent\ of\ solute }{volume \ of \ solution}}$