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Sever21 [200]
3 years ago
5

: How much energy is required to heat an iron nail with a mass of 25.5 grams from 65°C until it becomes red hot at 720°C?

Chemistry
1 answer:
Evgen [1.6K]3 years ago
8 0

Q = mct  

-Q= energy in Joules  

-m = mass in grams  

-c= specific heat capacity in J/g degree C  

-t = delta temperature in degrees Celsius  

So,  

Q = m c t  

Q = (7 grams)(0.448J/g C)(750 C - 25 C)  

Q = 2273.6 J  

Your final answer = 2273.6 Joules

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Can someone answer that first this pls im lost
Alchen [17]
-Just look up “H2O lewis structure
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4 0
3 years ago
A compound that is composed of molybdenum (Mo) and oxygen (O) was produced in a lab by heating molybdenum over a Bunsen burner.
irakobra [83]
First, we determine the mass of each element from the data collected. We can get the mass of molybdenum Mo from the difference between the mass of crucible and molybdenum and the mass of crucible:
     Mass of molybdenum = 39.52 – 38.26 = 1.26 g Mo

We can calculate for the mass of molybdenum oxide from the difference between the mass of crucible and molybdenum oxide and the mass of crucible:
     Mass of molybdenum oxide = 39.84 – 38.26 = 1.58g 

We can now compute for the mass of oxygen O by subtracting the mass of molybdenum from the mass of molybdenum oxide:
     Mass of oxygen in molybdenum oxide = 1.58 – 1.26 = 0.32g O

To convert mass to moles, we use the molar mass of each element.
     1.26 g Mo * 1 mol Mo / 95.94 g Mo = 0.0131 mol Mo
     0.32 g O * 1 mol O / 15.999 g O = 0.0200 mol O

0.0131 mol is the smallest number of moles. We divide each mole value by this number:
     0.0131 mol Mo / 0.0131 = 1
     0.0200 mol O / 0.0131 = 1.53

Multiplying these results by 2 to get the lowest whole number ratio,
     0.0131 mol Mo / 0.0131 = 1 * 2 = 2
     0.0200 mol O / 0.0131 = 1.5 * 2 = 3
Thus, we can write the empirical formula as Mo2O3.
5 0
3 years ago
Read 2 more answers
A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equiva
maxonik [38]

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of HNO_3

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of C_6H_5NH_2.

M_2\text{ and }V_2 are the final molarity and volume of HNO_3.

We are given:

M_1=0.3403M\\V_1=160.0mL\\M_2=0.0501M\\V_2=?

Putting values in above equation, we get:

0.3403M\times 160.0mL=0.0501M\times V_2\\\\V_2=1086.79mL

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of C_6H_5NH_2 +  Volume of HNO_3

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

\text{Concentration of salt}=\frac{0.3403M}{1246.79mL}\times 160.0mL=0.0437M

Now we have to calculate the pH of the solution.

At equivalence point,

pOH=\frac{1}{2}[pK_w+pK_b+\log C]

pOH=\frac{1}{2}[14+4.87+\log (0.0437)]

pOH=8.76

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-8.76\\\\pH=5.24

Thus, the pH of the solution is, 5.24

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