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AleksAgata [21]
2 years ago
5

How can you solve this ?

Chemistry
1 answer:
Karo-lina-s [1.5K]2 years ago
4 0
Math worksheet solved it
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Write the chemical equations for the neutralization reactions that occurred when hcl and naoh were added to the buffer solution.
irga5000 [103]
A base and an Acid always react to form a salt and water

So, HCl + NaOH —> NaCl + HOH
5 0
2 years ago
How many grams of CO are needed to produce 209.7 g Fe?<br><br> Fe2O3+3CO⟶3CO2+2Fe
kirill115 [55]

Answer:

157.8 g

Explanation:

Step 1: Write the balanced equation

Fe₂O₃ + 3 CO ⟶ 3 CO₂ + 2 Fe

Step 2: Calculate the moles corresponding to 209.7 g of Fe

The molar mass of Fe is 55.85 g/mol

209.7 g × 1 mol/55.85 g = 3.755 mol

Step 3: Calculate the moles of CO needed to produce 3.755 moles of Fe

The molar ratio of CO to Fe is 3:2. The moles of CO needed are 3/2 × 3.755 = 5.633 mol

Step 4: Calculate the mass corresponding to 5.633 moles of CO

The molar mass of CO is 28.01 g/mol.

5.633 mol × 28.01 g/mol = 157.8 g

4 0
2 years ago
What is the name of the supercontinent that existed 245 million years ago
Mandarinka [93]
I believe it was Pangaea but I could be wrong
6 0
3 years ago
Read 2 more answers
___________ have a charge of +1.
stiv31 [10]

Answer:

B) Protons

Explanation:

  1. A proton has positive charge of 1.
  2. Electrons are the negatively charged particles of atom.
  3. An atom consists of a positively charged nucleus surrounded by one or more negatively charged particles called electrons
  4. To find the ionic charge of an element you'll need to consult your Periodic Table. On the Periodic Table metals (found on the left of the table) will be positive. Non-metals (found on the right) will be negative.
3 0
3 years ago
Consider the following unbalanced reaction: P4(s) + F2(g) → PF3(g) What mass of fluorine gas is needed to produce 120. g of PF3
Studentka2010 [4]

Answer:

44.28 grams.

Explanation:

Let us write the balanced reaction:

P_{4}+6F_{2}-->4PF_{3}

As per balanced equation, six moles of fluorine gas will give four moles of PF₃.

The mass of PF₃ required = 120 g

The molar mass of PF₃ = 88g/mol

Moles of PF₃ required =\frac{mass}{molarmass}=\frac{120}{88}=1.364mol

The moles of fluorine gas required = \frac{4X1.364}{6}=0.91

the mass of fluorine gas required = moles X molar mass = 0.91x38 = 34.58g

Now this much mass will be required if the reaction is of 100% yield

But as given that the yield of reaction is only 78.1%

The mass of fluorine required = \frac{massX100}{78.1} =\frac{34.58X100}{78.1} =44.28g

4 0
3 years ago
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