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A water wave has a frequency of 10 Hz and a wavelength of 150 m. What is the wave speed? Equation to use: wave speed (m/s) = frequency (Hz) x wavelength (m)

Answers

Answer 1

Answer:

Answer:

Speed of wave = 1500 m/s

Explanation:

Given data:

Frequency of wave = 10 Hz

Wavelength of wave = 150 m

Speed of wave = ?

Solution:

Formula:

Speed of wave = Frequency × Wavelength

Now we will put the values in formula.

Speed of wave = 10 Hz × 150 m

Speed of wave = 1500 m/s

Hz = s⁻¹

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Which aqueous solution is colored?1.
CuSO4(aq)
2.
BaCl2(aq)
3.
KCl(aq)
4.
MgSO4(aq)

Answers

Answer: Copper (II) Sulfate! (CuSO4)

Explanation:

This is because compared to the other salts, it is the visible color blue. The other ones are plain white. Happy holidays! Don’t stress :)

How is Hess's law used to measure enthalpy of a desired reaction?A. The enthalpy is obtained from the enthalpy of an intermediate
step.
B. The enthalpy is determined from the enthalpy of similar reactions.
C. The enthalpy from the final equation in a series of reactions is
used
D. Intermediate equations with known enthalpies are added together.

Answers

Hess's law is used to measure the enthalpy of a desired chemical reaction because: D. Intermediate equations with known enthalpies are added together.

What is Hess's Law?

Hess's Law is also known as Hess's law of constant heat summation (enthalpy) and it was named after a Swiss-born Russian chemist called Germain Hess.

Hess's Law states that the energy change (enthalpy) experienced in a desired chemical reaction is equal to the sum of the energy changes (enthalpies) in each chemical reactions that it is made up of or contains.

Read more on Hess's Law here: brainly.com/question/9328637

100.0 g of liquid copper (molar mass 63.546 g/mol; melting point 1358 K; density 8.02 g/mL) is placed in a rigid container of volume 10.0 L at temperature 1508 K. The container is placed in an evacuated chamber and a small hole of area 3.23 mm2 is made in the upper container wall. After 2.00 hours, the mass of copper in the container has decreased by 1 0.0168 g. Assuming the mass loss is due to effusion, calculate the vapor pressure of liquid copper at 1508 K. Hint: because the liquid constantly evaporates, the pressure inside the container is constant

Answers

Answer:

8.912x10^-18

Explanation:

-dn/dt = pANa/2piMRT

100 g = initial copper

Number of moles = 100/63.546

= 1.5736

Mass of copper left = 100-10.0168

= 89.9832

Moles = 89.9832/63.546

= 1.4160

dn = 1.4160-1.5736

= -0.1576

dt = 2 hrs

A = 3.23mm² = 3.23x10^-6

M = 63.546

T = 0.0821

T = 1508k

Na = 6.023x10²³

When we insert all these into the formula above

We get

P = 8.912x10^-18atm

Transpiration is the loss of water from the leaves of plants. The stomata of leaves must open to allowcarbon dioxide to enter the leaf for photosynthesis, but when they are open, water vapor escapes into
the atmosphere.
HYPOTHESIS: As the intensity of light is increased, the rate of transpiration will increase, as
measured in by the loss of mass of the plant.
Independent variable:
Dependent variable:

Answers

Independent variable are the light intensity that what we change and dependent variable are the rate of transpiration that what we measure.

What is photosynthesis ?

Photosynthesis is the process that can be used by the plants and other animals to covert light energy into the chemical energy. Photosynthesis is the main source of food in the earth.

Photosynthesis is the process in which oxygen is released. For survival oxygen is very important, from this process we obtain sufficient amount of oxygen.

The dependent variable is the rate of transpiration and the independent variable is time.The dependent variable is the rate of transportation because it is depends on the environmental factor the plant is placed.

Thus,Independent variable are the light intensity that what we change and dependent variable are the rate of transpiration that what we measure.

To learn more about the photosynthesis, follow the link;

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#SPJ2

Answer:

Independent variable: The light intensity (what YOU CHANGE)

Dependent variable: The rate of transpiration (what YOU MEASURE)

Find the equilibrium constants, Kp, for the following equilibria, (i) NO(g) + ½ O2(g) ⇄ NO2(g), Kp = ? (ii) NO2(g) ⇄ NO(g) + ½ O2(g), Kp = ?, given the equilibrium constant, Kp, for the reaction: 2NO (g) + O2(g) ⇄ 2NO2(g) Kp= 100 at the same temperature

Answers

Answer :

(i) The value of equilibrium constants for this reaction is, 10

(ii) The value of equilibrium constants for this reaction is, 0.1

Explanation :

The given equilibrium reaction is,

$2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)$ $K_p=100$

Now we have to determine the equilibrium constants for the following equilibrium reactions.

(i) $NO(g)+(1)/(2)O_2(g)\rightleftharpoons NO_2(g)$ $K_p_1=?$

From the given reaction we conclude that, the reaction (i) will takes place when the given main reaction will be multiplied by half (1/2). That means when reaction will be half then the equilibrium constant will be:

$K_p_1=(K_p)^{(1)/(2)}$

$K_p_1=(100)^{(1)/(2)}$

$K_p_1=10$

The value of equilibrium constants for this reaction is, 10

(ii) $NO_2(g)\rightleftharpoons NO(g)+(1)/(2)O_2(g)$ $K_p_2=?$

From the given reaction we conclude that, the reaction (ii) will takes place when the reaction (i) will be reverse. That means when reaction will be reverse then the equilibrium constant will be:

$K_p_2=(1)/((K_p_1))$