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3 years ago

How are salinity, temperature and circulation related?

Chemistry

1 answer:

frosja888 [35]3 years ago

4 0

Answer:

Salinity, along with temperature, determines the density of seawater, and hence its vertical flow patterns in thermohaline circulation. 2. Salinity records the physical processes affecting a water mass when it was last at the surface. Hope this helps!

Explanation:

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hram777 [196]

Your answer would be c.

4 0

3 years ago

Copper and silver are examples that will form what bond

Helen [10]

A metallic bond since both are pure metals but are not ionic

6 0

3 years ago

What is the mass of an object with a net force of 10 N accelerating at 2 m/s^2??

xz_007 [3.2K]

Answer:

5kg

Explanation:

Force = Mass x acceleration

F = ma

m = F/a = 10N/2m/s^2

m = 10/2 = 5kg

The standard unit for mass = Kilogram

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3 years ago

Convert a speed of 259 cm/s to units of inches per minute. Also, show the unit analysis by dragging components into the unit‑fac

igomit [66]

Answer:

$6118.11\frac{in}{min}$

Explanation:

Hello,

In this case, by knowing that 1 inch equals 2.54 cm and 60 seconds equals 1 min, the resulting value results:

$=259\frac{cm}{s}*\frac{1in}{2.54cm}*\frac{60s}{1min}\\ \\=6118.11\frac{in}{min}$

Best regards.

5 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago