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3 years ago

What is the correct prefix for 5? A. Deca- B. Penta- C. Tetra- D. Hexa-

Chemistry

2 answers:

Art [367]3 years ago

6 0

Answer:

Its B penta

Explanation:

a_sh-v [17]3 years ago

5 0

Answer:

B. penta

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Calculate the average time the car took to reach each checkpoint. Record the average time in Table D of your Student Guide. The

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Answer:2.07

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4.92

Explanation:

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2 years ago

Read 2 more answers

A certain substance makes up 4.8 × 10–4 percent by mass of a normal healthy human being. How many grams of that substance would

iris [78.8K]

Answer:

0.2615 g

Explanation:

Given that mass of the body = 120 lb

The conversion of lb to g is shown below as:-

1 lb = 454 g

So, 120 lb = 120*454 g = 54480 g

The substance is $4.8\times 10^{-4}\ \%$ of the human body

So, Mass of that substance in the body of mass 54480 g is:-

$m=\frac{4.8\times 10^{-4}}{100}\times 54480\ g=0.2615\ g$

<u>0.2615 g of that substance would be found in the body of a person weighing 120 lb.</u>

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3 years ago

DOUBLE POINTS!! Please help ASAP!! 90 minutes left !! If the [H+] in a solution is 1×10^- 1 mol/L, what is the [OH-] ? PLEASE sh

ioda

Here is the answer, in the photo.

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2 years ago

27. The density of nickel is 8.91 g/cm3. How large a cube, in cm3, would contain 2.00 x 10^24 atoms of nickel? Use dimensional a

jok3333 [9.3K]

Answer : The volume of the cube is, $21.88cm^3$

Solution : Given,

Density of nickel = $8.91g/cm^3$

Number of nickel atoms = $2\times 10^{24}$

Molar mass of nickel = 58.7 g/mole

First we have to calculate the moles of nickel.

As, $6.022\times 10^{23}$ atoms form 1 mole of nickel

So, $2\times 10^{24}$ atoms form $\frac{2\times 10^{24}}{6.022\times 10^{23}}=3.321$ moles of nickel

The moles of nickel = 3.321 moles

Now we have to calculate the mass of nickel.

$\text{ Mass of Ni}=\text{ Moles of Ni}\times \text{ Molar mass of Ni}$

$\text{ Mass of Ni}=(3.321moles)\times (58.7g/mole)=194.94g$

The mass of nickel = 194.94 g

Now we have to calculate the volume of nickel.

$Density=\frac{Mass}{Volume}$

$8.91g/cm^3=\frac{194.94g}{Volume}$

$Volume=21.88cm^3$

Therefore, the volume of the cube is, $21.88cm^3$

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2 years ago

4 g of ag2so4 will dissolve in 1l of water. calculate the solubility product (ksp) for silver (i) sulfate.

Bas_tet [7]

Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.

7 0

3 years ago