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xenn [34]
3 years ago
10

An antacid tablet contains 640.0 mg of magnesium oxide per tablet.

Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
3 0

Answer:

317.6 mL

Explanation:

Step 1: Write the balanced neutralization equation

MgO + 2 HCl ⇒ MgCl₂ + H₂O

Step 2: Calculate the mass corresponding to 640.0 mg of MgO

The molar mass of MgO is 40.30 g/mol. The moles corresponding to 640.0 mg (0.6400 g) of MgO are:

0.6400 g × (1 mol/40.30 g) = 0.01588 mol

Step 3: Calculate the moles of HCl that react with 0.01588 moles of MgO

The molar ratio of MgO to HCl is 1:2. The moles of HCl are 2/1 × 0.01588 mol = 0.03176 mol

Step 4: Calculate the volume of 0.1000 M HCl that contains 0.03176 moles

0.03176 mol × (1 L/0.1000 mol) = 0.3176 L = 317.6 mL

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A chef is testing out a new set of iron cookware. He knows that iron has a specific heat of 445 J/kg°C. He heats a 2 kg iron ski
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5 0
3 years ago
The question in the picture, I really a correct answer, no cheap answers​
Delvig [45]

Answer:

94.325 g

Explanation:

We'll begin by converting 350 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

350 mL = 350 mL × 1 L /1000 mL

350 mL = 0.35 L

Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:

Volume = 0.35 L

Molarity of KC₂H₃O₂ = 2.75 M

Mole of KC₂H₃O₂ =?

Molarity = mole /Volume

2.75 = Mole of KC₂H₃O₂ / 0.35

Cross multiply

Mole of KC₂H₃O₂ = 2.75 × 0.35

Mole of KC₂H₃O₂ = 0.9625 mole

Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:

Mole of KC₂H₃O₂ = 0.9625 mole

Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)

= 39 + 24 + 3 + 32

= 98 g/mol

Mass of KC₂H₃O₂ =?

Mass = mole × molar mass

Mass of KC₂H₃O₂ = 0.9625 × 98

Mass of KC₂H₃O₂ = 94.325 g

Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g

6 0
2 years ago
if a sample of gas at 25.2 c has a volume of 536mL at 637 torr, what will its volume be if the pressure is increased to 712 torr
nignag [31]
Considering ideal gas:
PV= RTn

T= 25.2°C = 298.2 K

P1= 637 torr = 0.8382 atm

V1= 536 mL = 0.536 L

:. R=0.082 atm.L/K.mol

:. n= (P1V1)/(RT) = ((0.8382 atm) x (0.536 L))/
((0.082 atmL/Kmol) x (298.2K))

:. n= O.0184 mol

Then,
P2= 712 torr = 0.936842 atm

V2 = RTn/P2 = [(0.082atmL/
Kmol) x (298.2K) x (0.0184mol) ]/(0.936842atm)

:.V2 = 0.4796 L
OR
V2 = 479.6 ml

6 0
3 years ago
1
zloy xaker [14]

Answer:

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