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3 years ago

The density of copper is 8.92 g/cm3. What is the volume in L of a 230.14 mg sample of copper?

1 answer:

8 0

D=M/V so you plug in 8.92 for the density and 230.14 for the mass. then solve and get 25.8. you may need to do dimensional analysis with the different units of measures

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Differentiate between a precipitate and an aqueous solution

antiseptic1488 [7]

Answer:

A precipate is a solid while an aqueous solution is liquid.

Explanation:

A precipitate is a solid which separates after a chemical reaction occurs. It is the solid product of the reaction.

An aqueous solution is formed when a substance is dissolved in water.

6 0

3 years ago

when 10.00 g of phosphorus reacts with oxygen, it produces 17.77 g of a phosphorus oxide. This phosphorus oxide was found to hav

Alex_Xolod [135]

Answer:

Molecular formula = P₄O₆

Explanation:

P(s) + O₂(g)------------------------------------⇒ PₓOₙ (g)

10g (17.77-10)g 17.77g

10g 7.77g 17.77g (gramme ratio)

The molecular mass of Phosphorus (P) = 31g/mole

The molecular mass of Oxygen atom (O) = 16g/mole

Mole ratio is given by:

P : O

10/31 7.77/16

0.3226 : 0.4856 Mole ratio---------------------------- (1)

Divide (1) through by 0.3226

1 : 1.5-------------------------------------------- (2)

From (2), the empirical formula for Phosphorus oxide :

Empirical formula = P₁O₁.₅

= PO₁.₅

The molecular formula can be calculated from below:

Since the molecular formula is a multiple of the empirical formula we have

Molecular formula = (PO₁.₅)ₙ----------------------------------- (3)

Since we are given the molecular mass of the oxide formed, we have:

(PO₁.₅)ₙ = 220-----------------------------(4)

[31 + (16 x 1.5)] x n = 220

[31 + 24]n = 220

55n =220

n = 4

Substituting into (3), we have :

Molecular formula = (PO₁.₅)₄

= P₄O₆

8 0

3 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

3 years ago

Number 39! Please explain step by step!!!

ahrayia [7]

Answer:

The answer to your question is The honda civic hybrid weights 13181.8 N or 1318.2 kg

Explanation:

Data

Weight = 2900 lb

Weight = ? N

mass = ? kg

1 kg ------- 10 N

0.22lb ---- 1N

Use proportions to solve this problem

0.22 lb ------------------- 1 N

2900 lb ----------------- x

x = (2900 x 1) / 0.22

x = 2900 / 0.22

x = 13181.8 N

1 kg ----------------------- 10 N

x ----------------------- 13181.8 N

x = (13181.8 x 1)/10

x = 1318.2 kg

7 0

3 years ago

Choose the covalent compounds from the following choices Br2 MgS SO2 KF

Novosadov [1.4K]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

Less than 0.4 then it is Non Polar Covalent Bond

Between 0.4 and 1.7 then it is Polar Covalent Bond

Greater than 1.7 then it is Ionic

For Br₂;

E.N of Bromine = 2.96

________

E.N Difference 0.00 (Non Polar Covalent Bond)

For MgS;

E.N of Sulfur = 2.58

E.N of Magnesium = 1.31

________

E.N Difference 1.27 (Ionic Bond)

For SO₂;

E.N of Oxygen = 3.44

E.N of Sulfur = 2.58

________

E.N Difference 0.86 (Polar Covalent Bond)

For KF;

E.N of Fluorine = 3.98

E.N of Potassium = 0.82

________

E.N Difference 3.16 (Ionic Bond)

Result: The Bonds in Br₂ and SO₂ are Covalent in Nature.

3 0

3 years ago