All categories

3 years ago

Frank has a sample of steel that weighs 80 grams. If the density of his sample of steel is 8 g/cm3 ,what is the samples volume?

Chemistry

1 answer:

kirill115 [55]3 years ago

4 0

Answer: 10 cm3

Explanation:

Mass of sample of steel (m) = 80 grams

density of sample of steel (p) = 8 g/cm3 samples volume (v) = ?

Since density is obtained by dividing mass by volume, the density of the steel sample is expressed as:

density = mass / volume

p = m / v

make v the subject formula

v = m / p

v = 80 grams / 8 g/cm3

v = 10 cm3

Thus, the samples volume is 10 cm3

You might be interested in

Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbol

77julia77 [94]

Answer:

The Lewis structures are in image attached.

Explanation:

Lewis symbol is a representation of an element symbol along with its valence electrons around it in the form of dot(s).

Mono-atomic ions formed from the following :

(a) Cl

Chlorine's atomic number is 17 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

$Cl=1s^22s^22p^63s^23p^5$

$Cl^-=1s^22s^22p^63s^23p^6$

(b) Na

Sodium's atomic number is 11 in which only 1 electrons are present in its valence shell .So in order to gain noble gas stability it will loose 1 electron to completes its octet. In the Lewis symbol no dot shown as sodium has lost its 1 electron.

$Na=1s^22s^23p^63s^1$

$Na^+=1s^22s^23p^63s^0$

(c) Mg

Magnesium's atomic number is 12 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electrons to completes its octet.

In the Lewis symbol no dot shown as magnesium has lost its 2 electrons.

$Mg=1s^22s^23p^63s^2$

$Mg^{2+}=1s^22s^23p^63s^0$

(d)Ca

Calcium's atomic number is 20 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

$Ca= 1s^22s^22p^63s^23p^64s^2$

$Ca^{2+}=1s^22s^23p^6^23p^64s^0$

(e) K

Potassium's atomic number is 19 in which only 1 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 1 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 1 electron.

$K= 1s^22s^22p^63s^23p^64s^1$

$K^{+}=1s^22s^23p^6^23p^64s^0$

(f) Br

Bromine's atomic number is 35 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

$Br=1s^22s^22p^63s^23p^63d^{10}4s^24p^5$

$Br^-= 1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

(g) Sr

Strontium's atomic number is 38 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

$Sr=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2$

$Sr^{2+}=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^0$

(h) F

Florine's atomic number is 7 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet.

$F=1s^22s^22p^5$

$F^-=1s^22s^22p^6$

5 0

2 years ago

Can you pls tell me the word equations for all these equations

Leokris [45]

Answer:

Below

Explanation:

Balanced form;

$2C_6H_6 +15O_2 -> 12CO_2+6H_2O\\\\ Ca_3(PO_4)_2 +8C -> Ca_3P_2 +8CO\\\\2HNO_2+O_2 -> 2HNO_3\\\\Ca(OH)_2 + CO_2 -> CaC0_3+H_2O\\\\2K +Br_2 ->2KBr\\\\2NaOH+FeSO_4 -. Na_2SO_4 +Fe(OH)_2$

1.Benzene + Dioxygen = Carbon Dioxide + Water

2.Tricalcium phosphate +Carbon = Calcium phosphide + carbon monoxide

3.Nitrous acid react with oxygen to produce nitric acid.

4.This means that the carbon dioxide and limewater react to produce calcium carbonate and water.

5.Potassium react with bromine to produce potassium bromide

6. An aqueous solution of ferrous sulphate reacts with aqueous solution of sodium hydroxide to form a precipitate of ferrous hydroxide and sodium sulphate remains in the solution.

5 0

3 years ago

What is the net ionic equation for the reaction that occurs when aqueous copper(II) sulfate is added to excess 6-molar ammonia?a

Karolina [17]

Answer:

c. 2NH₃ + 2H₂O + Cu²⁺ → Cu(OH)₂(s) + 2NH₄⁺

Explanation:

A net ionic equation is a chemical equation that list only the species that are involved in the reaction.

The reaction of ammonia with copper(II) sulfate CuSO₄ in water is:

2NH₃ + 2H₂O + CuSO₄ → Cu(OH)₂(s) + 2NH₄⁺ + SO₄²⁻

In an ionic equation, salts are written as ions, that means CuSO₄ must be written as Cu²⁺ + SO₄²⁻. That is:

2NH₃ + 2H₂O + Cu²⁺ + SO₄²⁻ → Cu(OH)₂(s) + 2NH₄⁺ + SO₄²⁻

As in a net ionic equation you must list only the species involved in the reaction (The underlined species don't react), the net ionic equation is:

c. 2NH₃ + 2H₂O + Cu²⁺ → Cu(OH)₂(s) + 2NH₄⁺

I hope it helps!

7 0

3 years ago

A deficiency in thiamine (TPP) will directly decrease the activity of which of the following enzymes? a. Pyruvate carboxylase b.

Blababa [14]

Answer:

d. Transketolase

Explanation:

Transketolase utilize thiamine diphosphate (ThDP) as cofactor has reduced activity during the thiamine deficiency and results in the tissue damage. The levels of steady-state mRNA of transketolase in human lymphoblasts, fibroblasts and neuroblastoma cells are found to be lower in the thiamine-deficient cultures.

Thus, deficiency in TPP directly decrease the activity of enzyme, transketolase.

3 0

3 years ago

A laboratory experiment requires 4.8 l of a 2.5 m solution of sulfuric acid (h2so4), but the only available h2so4 is a 6.0 m sto

Airida [17]

To determine the amount of 6.0 M H2SO4 needed for the preparation, equate the number of moles of the 6.0 M and 2.5 M H2SO4 solution. This is done as follows

M1 x V1 = M2 x V2

Substituting the known variables,

(6.0 M) x V1 = (2.5 M) x (4.8 L)

Solving for V1 gives an answer of V1 = 2 L. Thus, to prepare the needed solution, dilute 2 L of 6.0 M H2SO4 solution with water until the volume reach 4.8 L.

5 0

3 years ago

Read 2 more answers