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3 years ago

How explain two ways to decrease the energy of a wave

Chemistry

1 answer:

Aleks04 [339]3 years ago

3 0

Answer:

gravity that's what I rellat think it is

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A 64.4-g sample of potassium chloride was added to 200.0 g of water at 20 degrees C. Is the solution saturated, unsaturated, or

Tomtit [17]

Hey there!:

is the solution <span>saturated , ie :

</span>They are the ones that have reached the exact solubility coefficient.<span>If we mix 64.4 g of KCl at 200.0 g of water at 20 º C, we'll see that the 32.2 g will dissolve and the remainder (32.2 g) will precipitate, forming the bottom body. In this case we will then have a solution saturated with background. However, if we want only the saturated solution, simply perform a simple filtration to separate the precipitate from the saturated solution.

hope this helps!</span>

7 0

3 years ago

The heat released by one mole of sugar from a bomb calorimeter experiment is 5648 kJ/mol. The balanced chemical reaction equatio

Bogdan [553]

Answer:

The answer to your question is the letter C) 5648 kJ/mol

Explanation:

Data

C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11 H₂O

H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol

H° O₂ = 0 kJ / mol

H° CO₂ = -393.5 kJ/mol

H° H₂O = -285.8 kJ/mol

Formula

ΔH° = ∑H° products - ∑H° reactants

Substitution

ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)

ΔH° = -4722 - 3143.8 + 2221.8

Result

ΔH° = -5644 kJ/mol

6 0

3 years ago

Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at

MariettaO [177]

Answer:

P_2 =0.51 atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}$

$0.008389261745 \times 273 = 4.5P_2$

$P_2 =\dfrac{0.008389261745 \times 273 }{4.5}$

$P_2 =0.51 \ atm$

4 0

2 years ago

18.2L of gas at 95°C and 760 torr is placed in a 15L container at 80 degrees * C ; what is the new pressure ?

romanna [79]

Answer:

884.56 torr

Explanation:

Formula: $\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}$

P = Pressure

V = Volume

T = Temperature in kelvin (Celsius + 273.15)

$\frac{(760)(18.2) }{368.15} = \frac{P(15) }{353.15}$

$P = \frac{(760)(18.2)(353.15) }{(368.15)(15)}$

P = 884.56169

4 0

3 years ago

The heat of combustion of ethane is -337.0kcal at 25 degrees celsius. what is the heat of the reaction when 3g of ethane is burn

Ghella [55]

Answer:

Q = -33.6kcal .

Explanation:

Hello there!

In this case, according to the equation for the calculation of the total heat of reaction when a fixed mass of a fuel like ethane is burnt, we can write:

$Q=n*\Delta _cH$

Whereas n stands for the moles and the other term for the enthalpy of combustion. Thus, for the required total heat of reaction, we first compute the moles of ethane in 3 g as shown below:

$n=3g*\frac{1mol}{30.08g}=0.1mol$

Next, we understand that -337.0kcal is the heat released by the combustion of 1 mole of ethane, therefore, to compute Q, we proceed as follows:

$Q=0.1mol*-337.0\frac{kcal}{mol}\\\\Q=-33.6kcal$

Best regards!

8 0

2 years ago