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sukhopar [10]
3 years ago
6

All metals are minerals a. true b. false​

Chemistry
1 answer:
Anettt [7]3 years ago
7 0

Answer:

B. False is the rite answer

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3 0
4 years ago
What are the two main types of research a scientist can pursue?
Jobisdone [24]

Answer:

biology nd geology those have to do with the earth

6 0
4 years ago
In a coffee-cup calorimeter, 150.0 mL of 0.50 M HCl is added to 50.0 mL of 1.00 M NaOH to make 200.0 g solution at an initial te
Zigmanuir [339]

Answer:

51.54°C the final temperature of the calorimeter contents.

Explanation:

HCl+NaOH\rightarrow H_2O+NaCl,\Delta H=-56 kJ/mol

moles=Molarity\times Volume (L)

Molarity of HCl= 0.50 M

Volume of HCl= 150.0 mL = 0.150 L

Moles of HCl= n

n=0.50 M\times 0.150 L=0.075 mol

Molarity of NaOH= 1.00 M

Volume of NaOH= 50.0 mL = 0.050 L

Moles of NaOH= n'

n'=1.00 M\times 0.050 L=0.050 mol

Since moles of NaOH are less than than moles of HCl. so energy release will be for neutralization of 0.050 moles of naOH by 0.050 moles of HCl.

n = 0.050

-56 kJ/mol=-\frac{Q}{n}

Q=56 kJ/mol\times 0.050 kJ/mol=2.8 kJ=2800 J

(1 kJ= 1000 J)

The energy change released during the reaction = 2800 J

Volume of solution = 150.0 mL + 50.0 mL = 200.0 mL

Density of the solution (water) = 1.00g/mL

Mass of the solution , m= 200 mL × 1.00 g/mL = 200 g

Now , calculate the final temperature by the solution from :

q=mc\times (T_{final}-T_{initial})

where,

q = heat gained = 2800 J

c = specific heat of solution = 4.184 J/^oC

T_{final} = final temperature = ?

T_{initial} = initial temperature = 48.2^oC

Now put all the given values in the above formula, we get:

2800 J=200.0 g\times 4.184 J/^oC\times (T_{final}-48.2)^oC

T_{final}= 51.54^oC

51.54°C the final temperature of the calorimeter contents.

7 0
3 years ago
Balance the following chemical equations by writing the appropriate coefficients in the blanks provided.
BartSMP [9]

Hey there!

Al + HCl → H₂ + AlCl₃

Balance Cl.

1 on the left, 3 on the right. Add a coefficient of 3 in front of HCl.

Al + 3HCl → H₂ + AlCl₃

Balance H.

3 on the left, 2 on the right. We have to start by multiplying everything else by 2.

2Al + 3HCl → 2H₂ + 2AlCl₃

Now we have 2 on the right and 4 on the left. Change the coefficient in front of HCl from 3 to 4.

2Al + 4HCl → 2H₂ + 2AlCl₃

Now, for Cl, we have 4 on the left and 6 on the right. Change the coefficient in front of HCl again from 4 to 6.

2Al + 6HCl → 2H₂ + 2AlCl₃

Now, our H is unbalanced again. 6 on the left, 4 on the right. Change the coefficient in front of H₂ from 2 to 3.

2Al + 6HCl → 3H₂ + 2AlCl₃  

Balance Al.

2 on the left, 2 on the right. Already balanced.

Here is our final balanced equation:

2Al + 6HCl → 3H₂ + 2AlCl₃  

Hope this helps!

5 0
3 years ago
A simple clear answer plz, Thanks.
Allushta [10]

Answer:

The number of energy levels increases as you move down a group as the number of electrons increases. Each subsequent energy level is further from the nucleus than the last. Therefore, the atomic radius increases as the group and energy levels increase. Hope this helps!

Explanation:

um same as the answer...?

4 0
3 years ago
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