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AVprozaik [17]
2 years ago
6

An object’s mass is determined to be 32.5 g. When the object is placed in a graduated cylinder containing 175 mL of water, the w

ater in the cylinder rises to a level of 187.5 mL. What is the density of the material?
Chemistry
1 answer:
MakcuM [25]2 years ago
6 0
<h3><u>Given :-</u></h3>
  • Mass of object = 32.5 g

<h3><u>To Find </u><u>:</u><u>-</u></h3>
  • Dentity of the material

<h3><u>Solution:- </u></h3>

When the object is placed in a graduated cylinder containing 175 ml of water, the water in the cylinder rises to a level of 187.5 ml , this implies that volume of content in the container has increased , the increased volume is volume of the object placed

So, Volume of object

= Increased Volume - Initial volume

= > 187.5 -175

= > 12.5 ml

Since 1 ml = 1 cm³ we can write

\quad12.5 ml = 12.5 cm³

Now,

\green{ \underline { \boxed{ \sf{Density=  \frac{Mass}{Volume}}}}}

Putting Values -

\begin{gathered}\\\implies\quad \sf Density =  \frac{32.5}{12.5}  \\\end{gathered}

\begin{gathered}\\\implies\quad \sf Density =  \frac{325}{125}  \\\end{gathered}

\begin{gathered}\\\implies\quad \sf Density = 2.6 g/cm^3 \\\end{gathered}

>> The density of the material is 2.6 g/cm³

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2C4H10+13O2--&gt;8CO2+10H2O Using the predicted and balanced equation, How many Liters of CO2 can be produced from 150 grams of
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Answer:  233 L of CO_2 will be produced from 150 grams of  C_4H_{10}

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}  \text{Moles of} C_4H_{10}=\frac{150g}{58g/mol}=2.59moles

The balanced chemical equation is:

2C_4H_{10}+13O_2(g)\rightarrow 8CO_2+10H_2O  

According to stoichiometry :

2 moles of C_4H_{10} produce =  8 moles of CO_2

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8 0
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3 years ago
Calculate the amount of heat in kilojoules required when 13.0 g of water at 52.5°C is converted to
Sedbober [7]

Answer:

Q = 10.10059 KJ

Explanation:

Given data:

Amount of water = 13.0 g

Initial temperature = 52.5°C

Final temperature = 238.2°C

Heat required to convert it into steam = ?

Solution:

Specific heat capacity of water is 4.184 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 238.2°C - 52.5°C

ΔT = 185.7 °C

by putting values,

Q = 13.0 g × 4.184 J/g.°C × 185.7 °C

Q = 10100.59 J

Jolue to Kj:

10100.59j × 1 KJ/1000 J

Q = 10.10059 KJ

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