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Doss [256]
3 years ago
11

A loaf of bread has a volume of 700 cm3 and a mass of 420 g. what is the density of the bread?

Chemistry
1 answer:
joja [24]3 years ago
8 0

We know that the equation for density is:

D=\frac{m}{V}

where D is the density, m is the mass in g, and V is the volume in either mL or cubic centimeters.

So then, given two of the variables, we can solve for density:

D=\frac{420g}{700cm^{3}}

D=\frac{0.6g}{cm^{3}}

Therefore, the density of the bread is 0.6g per cubic centimeter.

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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
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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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3 years ago
Kumar is producing the photoelectric effect by using red light. He wants to increase the energy of emitted electrons. Based on t
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He has to use a different colored light at a higher frequency.
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When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o
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Answer:

The net ionic equation for the given reaction :

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)..[2]

Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

NaI(aq)\rightarrow Na^+(aq)+I^-(aq)

Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

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Coefficients of the correct balanced equation for the reaction, CaCl2 + AgNO3 ⟶ AgCl + Ca(NO3)2.
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