Answer:
1.69 × 10² ball bearings
Explanation:
The output of a plant is 4335 pounds of ball bearings per week (five days). The output of the plant per day is:
4335 lb/5 day = 867.0 lb/day
1 pound is equal to 453.592 grams. The mass in grams made each day is:
867.0 lb/day × (453.592 g/1 lb) = 1.911 g/day
Each ball bearing weighs 0.0113 g. The number of ball bearings made each day is:
1.911 g/day × (1 ball bearing/0.0113 g) = 1.69 × 10² ball bearing
Answer:
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)
Explanation:
Which ONE of the following is an oxidation–reduction reaction?
A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.
B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.
C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.
D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.
Answer:
0.02498 L
Explanation:
CaCO₃ + 2 HCl ➔ CaCl₂ + H₂O + CO₂
You have to use stoichiometry. According to the chemical equation, for every one mole of CaCO₃, two moles of HCl is needed for the reaction to occur. Before you can use this relation you need to convert grams of CaCO₃ to moles. To convert, you need to use the molar mass.
Molar mass of CaCO₃ = 100.086 g/mol
(0.2500 g)/(100.086 g/mol) = 0.002498 mol CaCO₃
Now using the relation of 1 mol of CaCO₃ for every 2 mol of HCl, convert moles of CaCO₃ to moles of HCl.
(0.002498 mol CaCO₃) (2 mol HCl)/(1 mol CaCO₃) = 0.004996 mol HCl
Since the molarity of the solution of HCl is 0.200 M (mol/L), you have to divide the amount of moles needed by the molarity of the solution.
(0.004996 mol HCl)/(0.200 M) = 0.02498 L
You will need 0.02498 L to react with 0.2500 g of CaCO₃.
Answer:
D) 0.450 M
Explanation:
Given data
- Initial volume (V₁): 6.00 mL
- Initial concentration (C₁): 0.750 M
- Volume of water (VH₂O): 4.00 mL
- Final volume (V₂): V₁ + VH₂O = 6.00 mL + 4.00 mL = 10.0 mL
- Final concentration (C₂): ?
We can determine the final concentration using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁/V₂
C₂ = 0.750 M × 6.00 mL/10.0 mL
C₂ = 0.450 M
well yah .... homogeneous mixture has the same uniform appearance and composition throughout. Many homogeneous mixtures are commonly referred to as solutions.
while on the other hand ...A heterogeneous mixture consists of visibly different substances or phases. The three phases or states of matter are gas, liquid, and solid.
there are many differences between these two .... but fr a basic info .... i guess what i said would help .....