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boyakko [2]
3 years ago
5

What are the atomic numbers for sodium,iron and sulfur???

Chemistry
2 answers:
Crank3 years ago
7 0

Answer:

sodium:11

iron:26

sulfur:16

Explanation:

pls..give..brainlist...to the..above..person..thx..u

Nastasia [14]3 years ago
3 0

Answer: sodium-11   iron-26   sulfur-16

Explanation:

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HELP I NEED HELP ASAPP
Effectus [21]

Answer:

B = A/DH – C

Explanation:

From the question given above, we obtained:

A = D • H(B + C)

Thus, we can obtain B in terms of D, H, A and C by doing the following:

A = D • H(B + C)

A = DH(B + C)

Divide both side by DH

A/DH = B + C

Subtract C from both side

A/DH – C = B + C – C

A/DH – C = B

B = A/DH – C

4 0
3 years ago
34 atoms of carbon (C) react with 22 molecules of hydrogen gas (H2). How many molecules of methane (CH4) will be formed, and wha
kolbaska11 [484]

Answer:

11 molecules of CH4.

23 atoms of C is the leftover.

Explanation:

Hello!

In this case, for the formation of methane:

C+2H_2\rightarrow CH_4

We can see there is an excess of carbon based on their stoichiometry, because the needed amount of hydrogen gas molecules would be:

molecules _{H_2}=34atomC*\frac{2molec\ H_2}{1atomC} =68molec\ H_2

Thus, the formed molecules of methane are computed below:

molec\ CH_4=22molec\ H_2 *\frac{1molec\ CH_4}{2molmolec\ H_2} \\\\molec\ CH_4=11molec\ CH_4

In such a way, the leftover of carbon atoms are:

atoms \ C^{left over}=34-22molec\ H_2*\frac{1atoms C}{2molec\ H_2} \\\\atoms \ C^{left over}=23 atoms C

Best regards!

4 0
3 years ago
Match the following phases of stellar evolution to its characteristic.
Artist 52 [7]

Answer:

im the only answer your gonna get

Explanation:

7 0
2 years ago
1.1 L of nitrogen dioxide were produced in the reaction seen below. How
Veseljchak [2.6K]
22.4 since it's in STp
4 0
3 years ago
What volume (L) of 0.250 M HNO3 is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water ?
IRINA_888 [86]
<h3>Answer:</h3>

1.75 L HNO₃

<h3>Explanation:</h3>

We are given;

Molarity of HNO₃ as 0.250 M

Mass of NaOH as 17.5 g

Volume of NaOH = 350 mL

We are required to calculate the volume of 0.250 M

We are going to first write the balanced reaction:

NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

Then, we calculate the number of moles of NaOH

Moles = Mass ÷ Molar mass

Molar mass of NaOH = 39.997 g/mol

          = 17.5 g ÷ 39.997 g/mol

          = 0.4375 moles

We can now calculate the number of moles of HNO₃ using the mole ratio from the equation;

Mole ratio of NaOH to HNO₃ is 1 : 1

Therefore, if moles of NaOH are 0.4375 moles then;

Moles of HNO₃ will also be 0.4375 moles

We can now calculate the volume of HNO₃

Morality = Number of moles ÷ Volume

Thus;

Volume = Number of moles ÷ Molarity

             = 0.4375 moles ÷ 0.250 M

             = 1.75 L

Therefore, the volume of HNO₃ is 1.75 L

5 0
3 years ago
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