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3 years ago

When a baseball bat hits a baseball, what happens to the energy? completo ancur

Chemistry

1 answer:

professor190 [17]3 years ago

3 0

Answer: the baseball bat transfor its energy in the ball.

Explanation: you hit it with one and find out and tell me what you think what happen.

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What are chemical symbols and chemical formulas used for

lozanna [386]

Answer:

chemicals symbols are used for abbreviating the name of the element/chemical while chemicals formulas tell you how much of each element are in each chemical atom

Explanation:

4 0

3 years ago

If your end product is 1.5 moles of KMnO 4 how many moles of manganese oxide were used in the reaction? The equation for the pro

vovikov84 [41]

Answer:

1.5 moles

Explanation:

The equation of the reaction is given as:

2 MnO2 + 4 KOH + O2 --> 2KMnO 4 + 2KOH + H2

From the equation,

2 moles of MnO2 produces 2 moles of KMnO4

x moles of MnO2 would produce 1.5 moles of KMnO4

2 = 2

x = 1.5

Solving for x;

x = 1.5 * 2 / 2

x = 1.5 moles

4 0

3 years ago

se tienen 10 litros de nitrogeno gaseoso bajo una presion de 560 mm Hg. Determina el volumen del nitrogeno a una presion de 760

Dominik [7]

Answer:

working business city philadelphian

3 0

2 years ago

Calculate the molar mass of carbon tetrafluoride (CF4) in grams per mole, rounding to proper significant figures, if mc= 12.01 u

Nata [24]

Answer:

molar mass of carbon tetrafluoride (CF4) is

(12.01 × 1 ) + ( 4 × 19.00)

= 12.01 + 76

= 88.01u

= 88u

Hope this helps

3 0

3 years ago

Read 2 more answers

5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wav

torisob [31]

Answer: The range of wavelengths of light that can be used to cause given phenomenon is $8.953 \times 10^{21} m$ .

Explanation:

Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J

Formula used is as follows.

$E = \frac{hc}{\lambda}$

where,

E = energy

h = Planck's constant = $6.625 \times 10^{-25} Js$

c = speed of light = $3 \times 10^{8} m/s$

Substitute the values into above formula as follows.

$E = \frac{hc}{\lambda}\\222000 J = \frac{6.625 \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m$

Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is $8.953 \times 10^{21} m$ .

7 0

3 years ago