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Luba_88 [7]
2 years ago
5

What is the density of a substance that has a volume of 14 millimeters and a mass of 13.2 grams.

Chemistry
1 answer:
Contact [7]2 years ago
6 0

Answer:

<h2>0.94 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

density =  \frac{13.2}{14}  \\  = 0.94285...

We have the final answer as

<h3>0.94 g/mL</h3>

Hope this helps you

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Which phase change in an example of an exothermic process?
photoshop1234 [79]
C. liquid to solid  i hope this helps

5 0
3 years ago
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When 1.045 g of CaO is added to 50.0 mL of water at 25.0 °C in a calorimeter, the temperature of the water increases to 32.3 °C.
Rus_ich [418]

Answer:

1.71 kJ/mol

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

\Delta H=m\times C\times \Delta T

Where,  

\Delta H  is the enthalpy change

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass of CaO = 1.045 g

Specific heat = 4.18 J/g°C

\Delta T=32.3-25.0\ ^0C=7.3\ ^0C

So,  

\Delta H=1.045\times 4.18\times 7.3\ J=31.88713\ J

Also, 1 J = 0.001 kJ

So,  

\Delta H=0.03189\ kJ

Also, Molar mass of CaO = 56.0774 g/mol

Moles=\frac{Mass}{Molar\ mass}=\frac{1.045}{56.0774}\ mol=0.01863\ mol

Thus, Enthalpy change in kJ/mol is:-

\Delta H=\frac{0.03189}{0.01863}\ kJ/mol=1.71\ kJ/mol

8 0
2 years ago
The movement of organisms into a range is called
Brrunno [24]
A. immigration, how is this chemistry?
8 0
2 years ago
NH3(aq) + HNO3(aq) → NH4NO3(aq) Calculate the volume of an acid (1.5 M HNO3) needed to neutralize the 1.5 M HNO3.
SCORPION-xisa [38]

Answer:

NH3(aq) + HNO3(aq) → NH4NO3(aq) Calculate the volume of an acid (1.5 M HNO3) needed to neutralize the 1.5 M HNO3.

Explanation:

8 0
2 years ago
A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The
natima [27]
1) Compund Ir (x) O(y)

2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

5) Convert grams to moles

moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

6) Find the proportion of moles

Divide by the least of the number of moles, i.e. 0.00656

Ir: 0.00656 / 0.00656 = 1

O: 0.0131 / 0.00656 = 2

=> Empirical formula = Ir O2 (where 2 is the superscript for O)

Answer: Ir O2
5 0
3 years ago
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