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Valentin [98]
3 years ago
8

Elements in the same column of the periodic table share:

Chemistry
1 answer:
avanturin [10]3 years ago
7 0
B. the same number of protons.
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Calculate the mass of liquid with density of 3.3 g/mL and a volume of 25 oz
love history [14]

Answer: 215

Explanation:

Volume doesn’t add up into ounces that’s mass only

3 0
3 years ago
How many moles of electrons must be transferred to plate out 110 g of manganese (MW ~ 55g/mol) from a solution of permanganate (
ololo11 [35]

Answer:

14 mol e⁻

Explanation:

Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese

8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)

Step 2: Calculate the moles corresponding to 110 g of manganese

The molar mass of Mn is 55 g/mol.

110 g × 1 mol/55 g = 2 mol

Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn

According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.

2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻

8 0
3 years ago
40. Write a balanced equation and a net ionic equation for combining AgNO3(aq) and Na2CO3(aq)
Kay [80]

we are given the the two reactants: AgNO3 and Na2CO3 and is asked to write a balanced equation and a net ionic equation for the reaction of the two. This is a double-replacement reaction:
2AgNO3 (aq)+ Na2CO3 (aq)= Ag2CO3 + 2NaNo3 (aq)
2 Ag + + 2 N03- + 2Na+ + CO32- = Ag2CO3 + 2 Na+ 2NO3-
cancelling the spectator ions, 2Ag + + CO32- = Ag2CO3 
3 0
3 years ago
Fill in the coefficients to balance the equation for the chemical reaction that occurs:
Dominik [7]

Answer:

The equation is balanced

Explanation:

NaCl (aq) + AgNO3(aq) ––> AgCl (s) + NaNO3 (aq)

NaCl (aq) + AgNO3 (aq)

Na = 1 , Cl=1 , Ag = 1 , No3= 1

AgCl (s) + NaNO3 (aq)

Ag = 1 , Cl=1 , Na = 1 , No3= 1

8 0
2 years ago
A chemical reaction can theoretically produce 137.5 grams of product, but in actuality 112.9 grams are
Butoxors [25]
<h3>Answer:</h3>

82.11%

<h3>Explanation:</h3>

We are given;

  • Theoretical mass of the product is 137.5 g
  • Actual mass of the product is 112.9 g

We are supposed to calculate the percentage yield

  • We need to know how percentage yield is calculated;
  • To calculate the percentage yield we get the ratio of the actual mass to theoretical mass and express it as a percentage.

Thus;

% yield = (Actual mass ÷ Experimental mass) × 100%

            = (112.9 g ÷ 137.5 g) × 100%

            = 82.11%

Therefore, the percentage yield of the product is 82.11 %

7 0
3 years ago
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