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Free_Kalibri [48]
2 years ago
10

Which factors cause a rising tide? A. The moon's gravity and Earth's rotation B. Density differences due to salinity and tempera

ture C. Cold and warm ocean currents D. The ocean's conveyor belt and gyres
Chemistry
2 answers:
icang [17]2 years ago
5 0
The answer is A)The moon's gravity and Earth's rotation. -Apex
Irina18 [472]2 years ago
3 0
It a because the moons gravity pulls the water and the earths rotation moves the bump/wave
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The relationship between molecular velocities and temperature is a _____ relationship.
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The correct answer that would best complete the given statement above would be option 2. <span>The relationship between molecular velocities and temperature is a direct relationship. In other words, their relationship is directly proportional. Hope that this is the answer that you are looking for. </span>
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One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF7). How many moles of CsXeF7 can be produced from t
larisa [96]
The balanced chemical equation is written as:

<span>CsF(s) + XeF6(s) ------> CsXeF7(s)

We are given the amount of </span>cesium fluoride and <span>xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride.  We calculate as follows:

11.0 mol CsF ( 1 mol </span>CsXeF7 / 1 mol CsF ) = 11.0 mol <span>CsXeF7</span>
6 0
3 years ago
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How much heat (in Joules) is needed to raise the temperature of 257g of ethanol (cethanol=2.4 J/g°C) by 49.1°C?
Sonbull [250]

Answer:

Q = 30284.88 j

Explanation:

Given data:

Mass of ethanol = 257 g

Cp = 2.4 j/g.°C

Chnage in temperature = ΔT = 49.1°C

Heat required = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values in formula.

Q = 257 g× 2.4 j/g.°C × 49.1 °C

Q = 30284.88 j

8 0
2 years ago
Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0
olya-2409 [2.1K]

Answer:

pH=-1.37

Explanation:

We are given that 25 mL of 0.10 M CH_3COOH is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of CH_3COOH=25mL=0.025 L

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL=\frac{35}{1000}=0.035 L

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of CH_3COOH=0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+=0.10\times 0.025=0.0025 moles

Number of moles of OH^-=0.10\times 0.01=0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+=\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L

pH=-log [H+]=-log [4.28\times 10^{-2}]=-log4.28+2 log 10=-0.631+2

pH=-1.37

6 0
3 years ago
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