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tatiyna
3 years ago
8

Calculate the speed of a dog running through a field if he is covering 23.7 meters in 54 seconds.

Chemistry
2 answers:
const2013 [10]3 years ago
7 0
0.43888888≅0.439
Explanation:
speed = distance/time
speed= 23.7/54
speed= 0.43888
≅0.439
Elenna [48]3 years ago
4 0

Explanation:

speed = distance/time

= 23.7/54 m/s

= 0.44 m/s

speed of a dog running through a field = 0.44 m/s

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PLEASE HELP AND THANKS
Kamila [148]

Answer:

torque =force*distance

4 0
3 years ago
A sample of gold has a mass of 15.7 g and displaces 0.81 cm3 of eater in a graduated cylinder. What is the density of gold?
Ostrovityanka [42]

Answer: 19.4 g/cm3

Explanation: density is the relationship between mass over volume.

So density of gold is 15.7g/0.81cm3 = 19.4 g/cm3

3 0
4 years ago
Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0
3 years ago
What do the sweat collection pads tell the scientist?
spayn [35]
A
Is the correct answer

I’m 95% sure
4 0
3 years ago
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