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maria [59]
2 years ago
15

6. An average Mastiff puppy weighs 2.72 kilograms. How many pounds is an average Mastiff puppy? (1lb = 453.6 g)

Chemistry
1 answer:
Sloan [31]2 years ago
4 0

The number of pounds present in 2.72 kg of Mastiff puppy is 5.996 pounds

<h3>What is weight?</h3>

Weight can be defined as the gravitation pull on a body. or it can also be defined as the product of the mass of a body and its acceleration due to gravity.

<h3>Units of weight</h3>

weight has several units and they include

  1. Newton
  2. pounds
  3. ounces
  4. Tons.

These units can be converted.

From the question,

If

  • 1 lb = 453.6 g

But,

  • 1 kg = 1000 g

Then,

  • 2.72 kg = 2720 g

Therefore,

  • 2.72 kg of Mastiff = (2720/453.6) pounds of average Mastiff = 5.996 pounds

Hence, The number of pounds present in 2.72 kg of Mastiff puppy is 5.996 pounds.

Learn more about weight here: brainly.com/question/2337612

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How many grams of NaOH are<br> needed to make 400. mL of<br> 1.50 M solution?
juin [17]

Answer:

24g of NaOH are required

Explanation:

Molarity, M, is an unit of concentration widely used in chemistry defined as the ratio between moles of solute (In this case, NaOH), and volume of solution in liters.

We can find the moles of NaOH and its mass with the volume and desired concentration as follows:

<em>Moles NaOH:</em>

400.0mL = 0.400L * (1.50mol / L) = 0.600 moles NaOH

<em>Mass NaOH -Molar mass: 40.0g/mol-:</em>

0.600 moles * (40.0g / mol) =

<h3>24g of NaOH are required</h3>
7 0
3 years ago
Which of these is an example of a physical change ?
Gre4nikov [31]
It's 1. Melting a substance. The rest are chemical changes
8 0
3 years ago
Which of the following solutions is more concentrated?<br> 0.50M KCl or 5.0% (w/v) KCl
HACTEHA [7]

.50 M KCl because 5% is the same as .05, which makes the .50M more concentrated.

6 0
3 years ago
How many seconds would it take to deposit 17.3 g of ag (atomic mass = 107.87) from a solution of agno3 using a current of 10.00
Brums [2.3K]
Data Given:

Time = t = ?

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 107.86/1 = 107.86 g

Amount Deposited = W = 17.3 g

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Solving for t,

                                          t  =  W F / I e
Putting values,
                                          t  =  (17.3 g × 96500) ÷ (10 A × 107.86 g)

                                          t  =  1547.79 s

                                          t  = 1.54 × 10³ s
5 0
3 years ago
| A solution containing 4.48 ppm KMnO4 exhibits
Artist 52 [7]

Answer:

Molar absorptivity or molar extinction co-effecient = 2120.14 cm⁻¹M⁻¹

Explanation:

First convert Concentration from ppm inM or mol/l

⇒ Molar mass of KMnO₄ = 158.03 g

⇒ 4.48 ppm = 4.48 mg/l = 4.48 x 10⁻³ g/l

⇒ Molarity = \frac{4.48 X10^{-3} }{158.03X 1(lit)} = 2.83 x 10⁻⁵ molar

Absorbance (A) = - log(T)     ( T = % transmittance)

                          = - log(0.859)

                          = 0.06

According to Lambert Beer's law

     

                 ε = \frac{A}{C X l}

      or,      ε = \frac{0.06}{2.83 X 10^{-5}X1 cm }

      or,      ε = 2120.14 cm⁻¹M⁻¹

Where

    ε = Molar absorptivity

    A = absorbance

    C = Molar concentration of KMnO₄ solution

     l = length  

6 0
3 years ago
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