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Doss [256]
4 years ago
7

Two 0.75 Amp loads are connected in parallel with a 2500 Milliamperehour Ni-Cd battery. Approximately how long can the battery p

rovide power to the load?
Chemistry
1 answer:
nikdorinn [45]4 years ago
4 0

Explanation:

It is given that two loads have 0.75 Ampere current each. And, they contain 2500 milli ampere per hour Ni-Cd battery.

As both the loads are connected in parallel. Hence, total current will be calculated as follows.

               I = I_{1} + I_{2}

                 = 0.75 A + 0.75 A

                 = 1.5 A

                 = 1.5 A \times \frac{1000 mA}{1 A}

                 = 1500 mA

Relation between time and capacity of battery is as follows.

             Capacity = Current × time (in hour)

therefore,        time = \frac{Capacity}{Current}

                                = \frac{2500 mA. h}{1500 A}

                                = 1.667 hr

Thus, we can conclude that the battery provide power to the load up to 1.667 hours.

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I think the one element that could provide one atom to making an ionic bond will be Sodium 

3 0
4 years ago
A buffer solution contains a weak acid, HA , and its conjugate base, A − . The buffer solution has a pH of 5.76 , and the weak a
ludmilkaskok [199]

Answer:

The relationship is expressed as follows: K_{a} = \frac{[H+][A-]}{[HA]}

Explanation:

Most acidic substances are weak acids and are therefore only partially ionized in acqeous solution. We cab use the equilibrium constant for the ionization of acid to express the extent to which the weak acid ionizes. If we represent a general weak acid as HA, we can write the equation for its ionization reaction like this:

K_{a} = \frac{[H+][A-]}{[HA]}

To calculate the pH of a weak acid, we use the equilibrium concentration of the reacted species and product.

Take for example:

HA → H + A⁻

where A id the conjugate base.

Knowing that x amount of acid reacts, we can solve like this:

HA → H + A⁻

H+ = antilog (pH)

thus, the pH of the acid is equals to H+ (initial) - H+ (equilibrium) ≈ H+ (initial)

4 0
4 years ago
A glass of room temperature water is placed into the refrigerator.Which of the following statements are true?
Liono4ka [1.6K]
The answer would be D or c

8 0
4 years ago
Which of the following quantum number combinations is not allowed in a ground-state atom?
ollegr [7]
The third option. The angular quantum number (l) can be any integer between 0 and n-1. Therefore, for the third option, l can be 0, 1, 2, 3, or 4—but not 5.
7 0
3 years ago
Na2CO3 + 2HCl ---------> 2NaCl + CO2 + H2O How many moles of NaCl are produced from the reaction of 1.67 x 1022 molecules of
dexar [7]

Answer:

0.0554 moles of NaCl are produced from the reaction of 1.67*10²² molecules of Na₂CO₃ with excess HCl.

Explanation:

The balanced reaction is:

Na₂CO₃ + 2 HCl → 2 NaCl + CO₂ + H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • HCl: 2 moles
  • NaCl: 2 moles
  • CO₂: 1 mole
  • H₂O: 1 mole

On the other hand, Avogadro's Number is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number applies to any substance.

In this case, you can apply the following rule of three: if 6.023*10²³ molecules of Na₂CO₃ are contained in 1 mole, 1.67*10²² molecules will be contained in how many moles?

amount of moles=\frac{1.67*10^{22}molecules*1mole }{6.023*10^{23}molecules}

amount of moles= 0.0277 moles

In this case, you can apply the following rule of three: if by stoichiometry 1 mole of Na₂CO₃ produces 2 moles of NaCl, 0.0277 moles of Na₂CO₃ will produce how many moles of NaCl?

amount of moles of NaCl=\frac{0.0277 moles of Na_{2} CO_{3}*2 moles of NaCl }{1 mole of Na_{2} CO_{3}}

amount of moles of NaCl= 0.0554 moles

<u><em>0.0554 moles of NaCl are produced from the reaction of 1.67*10²² molecules of Na₂CO₃ with excess HCl.</em></u>

8 0
3 years ago
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