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bezimeni [28]
2 years ago
15

If a 60-g object has a volume of 30 cm3, what is its density? 2 g/cm3 0.5 cm3/g 1800 g * cm3 none of the above

Chemistry
1 answer:
jasenka [17]2 years ago
6 0

Answer : The density of an object is, 2g/cm^3

Solution : Given,

Mass of an object = 60 g

Volume of an object = 30cm^3

Formula used :

\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}

Now put all the given values in this formula, we get the density of an object.

\text{Density of an object}=\frac{60g}{30cm^3}=2g/cm^3

Therefore, the density of an object is, 2g/cm^3

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The value of resistance r was determined by measuring current I flowing through the resistance with an error E1=±1.5% and power
frozen [14]

Answer:5

Explanation:help me to solve

5 0
2 years ago
In the electrochemical cell using the redox reaction below, the anode half reaction is ________. Sn4+ (aq) + Fe (s) → Sn2+ (aq)
kenny6666 [7]

Answer:

The anode half reaction is : Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}

Explanation:

In electrochemical cell, oxidation occurs in anode and reduction occurs in cathode.

In oxidation, electrons are being released by a species. In reduction, electrons are being consumed by a species.

We can split the given cell reaction into two half-cell reaction such as-

Oxidation (anode): Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}

Reduction (cathode): Sn^{4+}(aq.)+2e^{-}\rightarrow Sn^{2+}(aq.)

------------------------------------------------------------------------------------------------------------

overall: Fe(s)+Sn^{4+}(aq.)\rightarrow Fe^{2+}(aq.)+Sn^{2+}(aq.)

So the anode half reaction is : Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}

8 0
2 years ago
If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, h
frutty [35]

Answer:

There were 0.00735 moles Pb^2+ in the solution

Explanation:

Step 1: Data given

Volume of the KI solution = 73.5 mL = 0.0735 L

Molarity of the KI solution = 0.200 M

Step 2: The balanced equation

2KI + Pb2+ → PbI2 + 2K+

Step 3: Calculate moles KI

moles = Molarity * volume

moles KI = 0.200M * 0.0735L = 0.0147 moles KI

Ste p 4: Calculate moles Pb^2+

For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+

For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

There were 0.00735 moles Pb^2+ in the solution

3 0
2 years ago
Can 750 mL of water dissolve 0.60 mol of gold(III) chloride (AuCl3)?
nydimaria [60]

Answer:

Yes  

Explanation:

1. Mass of 0.60 mol of AuCl₃  

\text{Mass} = \text{0.60 mol} \times \dfrac{\text{303.33 g}}{\text{1 mol}} = \text{184 g}

2. Mass of AuCl₃ in 750 mL

The solubility of AuCl₃ is 68 g/100 mL.

In 750 mL of water, you can dissolve

\text{Mass of AuCl}_{3} = \text{750 mL} \times \dfrac{\text{68 g}}{\text{100 mL}} = \text{510 g AlCl}_{3}

∴ Yes, 750 mL of water can dissolve 0.60 mol of AuCl₃.

3 0
2 years ago
Which of the following atoms would have the longest de Broglie wavelength, if all have the same velocity?
GenaCL600 [577]

Answer:

Li

Explanation:

The phenomenon of wave particle duality was well established by Louis deBroglie. The wavelength associated with matter waves was related to its mass and velocity as shown below;

λ= h/mv

Where;

λ= wavelength of matter waves

m= mass of the particle

v= velocity of the particle

This implies that if the velocities of all particles are the same, the wavelength of matter waves will now depend on the mass of the particle. Hence; the wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum. The longest wavelength will then be obtained from the smallest mass of matter. Hence lithium which has the smallest mass will exhibit the longest DeBroglie wavelength

4 0
2 years ago
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