I don’t know how to answer these, can anyone answer any of them?

Chemistry

1 answer:

vlabodo [156]4 years ago

3 0

Answer:

1.) To convert between grams and moles, you would use the substance's molar mass. To go from grams to moles, divide the grams by the molar mass. 600 g58.443 g/mol = 10.27 mol of NaCl. It has been found that 1 mol of any gas at STP (Standard Temperature and Pressure = 0 °C and 1 atm) occupies 22.4 L

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(a) show that the pressure exerted by a fluid P (in pascals) is given by P= hdg, where h is the column of the fluid in metres, d

Murrr4er [49]

Answer:

56.4 m

Explanation:

volume increases by factor of 6, i.e $\frac{V2}{V1}$ = 6

Initial temperature T1 at bottom of lake = 5.24°C = 278.24 K

Final temperature T2 at top of lake = 18.73°C = 291.73 K

NB to change temperature from °C to K we add 273

Final pressure P2 at the top of the lake = 0.973 atm

Initial pressure P1 at bottom of lake = ?

Using the equation of an ideal gas

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

P1 = $\frac{P2V2T1}{V1T2}$ = $\frac{0.973*6*278.24}{291.73}$

P1 = 5.57 atm

5.57 atm = 5.57 x 101325 = 564380.25 Pa

Density Ρ of lake = 1.02 g/ $cm^{3}$ = 1020 kg/ $m^{3}$

acceleration due to gravity g = 9.81 $m/s^{2}$

Pressure at lake bottom = pgd

where d is the depth of the lake

564380.25 = 1020 x 9.81 x d

d = $\frac{564380.25}{10006.2}$ = 56.4 m

3 0

4 years ago

WARRIOR [948]

Answer:

Explanation:

3 0

3 years ago

Read 2 more answers

For the reaction N2(g) + O2(g)2NO(g) H° = 181 kJ and S° = 24.9 J/K G° would be negative at temperatures (above, below) K. Enter

12345 [234]

ΔG deg will be negative above 7.27e+3 K.

Explanation:

The ΔG deg with the temperature can be found using the formula and the formula is given below
ΔG deg = ΔH deg - T ΔS deg
Given data, ΔH deg = 181kJ and ΔSdeg=24.9J/K
-T ΔS deg will be always negative and ΔG deg = ΔH deg will be positive and ΔG deg will be negative at relatively high temperatures and positive at relatively low temperatures
solving the equation and substitute ΔGdeg=0
ΔGdeg = ΔHdeg - T ΔSdeg
T= ΔHdeg/ΔSdeg
T=181 kJ / 2.49e-2 kJK-1
By simplification we get
T=7.27 × 10^3 K.
Therefore, Go will be negative above 7.27 × 10^3 K
Since ΔG deg = -RT lnK, when ΔGdeg < 0, K > 1 so the reaction will have K > 1 above 7.27 × 10^3 K.