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castortr0y [4]
3 years ago
11

As the temperature of seawater increases from 15o C to 20o C, sound velocity through the seawater

Chemistry
2 answers:
Thepotemich [5.8K]3 years ago
8 0

Answer: D- increases

Explanation:

polet [3.4K]3 years ago
7 0
Answer is: d. increases.
<span>The velocity of sound in seawater changes with water pressure, depth, temperature and salinity.
</span>Velocity can be calculeted using Del Grosso's equation, the UNESCO equation (<span>International Standard algorithm)</span> or <span>Mackenzie equation.
To measure velocity navy use </span>Expendable Bathy Thermograph.

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How many moles of CO2 would be present in a gas sample of 10 L at 25.0oC and a pressure of .77 atm?
Step2247 [10]
Use PV=nRT to solve the equation. You need to solve for n (number of moles). Don’t forget to convert the temperature to kelvins by adding 25+273. Use 0.082057 for R.
4 0
3 years ago
The smallest particle in any element is a compound. *
zubka84 [21]

Answer:

false.

Explanation:

the smallest particle of a element is an atom

7 0
3 years ago
Consider the following equilibrium: H2CO3+H2O = H3O+HCO3^-1. What is the correct equilibrium expression?
mylen [45]

Equilibrium expression is Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

<u>Explanation:</u>

Equilibrium expression is denoted by Keq.

Keq is  the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

Example -

aA + bB = cC + dD

So, Keq = conc of product/ conc of reactant

Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}

So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹

Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}

The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.

Thus,

Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

Therefore, Equilibrium expression is Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\

4 0
3 years ago
The diagram below shows a cell placed in a solution. What will happen to the cell?
drek231 [11]

the cell will expand as water moves into it.

7 0
3 years ago
Suppose of potassium sulfate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium
dimulka [17.4K]

Answer:

This question is incomplete, here's the complete question:

<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>

Explanation:

Reaction :-

K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4

Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol

Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol

Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L

Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L

Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol

Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.

0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+

Final concentration of potassium cation

= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M

8 0
3 years ago
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