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anastassius [24]
2 years ago
6

The stock concentration of NaOH is 0.5M. 0.5M = 0.500mol/1.0L.

Chemistry
1 answer:
valentinak56 [21]2 years ago
5 0

Answer:a) 0.1 mole. b) 4g. c) 2% d)  196 mL

Explanation: in 200mL , 0.1mole

mw NaOH = 40g/mol —> 4g in 0.1 mole

4g in 200mL so 2g in 100mL

density NaOH = 1g/mL so if 4g in 200 mL,  4mL , 196 mL water

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Explain why beryllium looses electrons while ionic bonds,while Sulfur gains electrons.
balandron [24]

Answer:

Since Beryllium has a larger atomic radius than Sulphur its electrons are not strongly attracted to the nucleus hence lost easily. But Sulphur has a small atomic radius hence electrons are more closely attracted to the nucleus.

3 0
2 years ago
The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C
artcher [175]

Explanation:

The given data is as follows.

         P_{1} = 100 mm Hg or \frac{100}{760}atm = 0.13157 atm

         T_{1} = 1080 ^{o}C = (1080 + 273) K = 1357 K

         T_{2} = 1220 ^{o}C = (1220 + 273) K = 1493 K

         P_{2} = 600 mm Hg or \frac{600}{760}atm = 0.7895 atm

          R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

                   log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}

            log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]

          log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]

                0.77815 = \frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K

              \Delta H_{vap} = 2.219 \times 10^{5} J/mol

                                   = 2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0
3 years ago
What type of reaction provide energy within a red giant star?<br><br> and how come?
Dafna1 [17]
"A red giant is a star that has exhausted the primary supply of hydrogen fuel at its core. An average-sized star like our Sun will spend the final 10 percent of its life as a red giant. In this phase, a star's surface temperature drops to between 3,140 and 6,741°F (1,727 and 3,727°C) and its diameter expands to 10 to 1,000 times that of the Sun. The star takes on a reddish color, which is what gives it its name."
3 0
3 years ago
An FAS calibration curve (absorbance, y-axis vs. FAS in g/mL) gave an equation he regression line of: y-3678(x)+0.056. If an unk
mezya [45]

Answer:

FAS concentration = 1.61*10^-4M

Explanation:

Beer Lambert's law relates the absorbance (A) of a substance to its concentration (c) as:

A = \epsilon lc----(1)

where ε = molar absorption coefficient

l = path length

A plot of 'A' vs 'c' gives a straight line with slope = εl

In addition absorbance (A) is related to % Transmittance (%T) as:

A = 2-log%T----(2)

For the FAS solution, the corresponding calibration fit is given as:

y = 3678(x) + 0.056

This implies that the slope = εl = 3678

It is given that %T = 25.6%

A = 2-log(25.6)=0.592

Based on equation(1):

c = \frac{A}{\epsilon l}=\frac{0.592}{3678}=1.61*10^{-4}M

8 0
3 years ago
Number 14? Please :)
ASHA 777 [7]
Evaporated is the answer
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