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Charra [1.4K]
2 years ago
13

Which of the following statements is a true statement regarding a solution with [H1+] =1x10-5 M and [OH1-]= 1x10-9 M?

Chemistry
1 answer:
zheka24 [161]2 years ago
4 0

Answer:

B. The [H1+] >[OH1-] and the solution is acidic

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What will happen to the volume if the number of moles of a gas is decreased at constant temperature and pressure?
RoseWind [281]
The moles will probably die
4 0
3 years ago
Read 2 more answers
If 5 moles of P4 reacted with 22 moles Cl2 according to the above reaction, determine:
GaryK [48]

Solution:

P_4+6Cl_2\rightarrow 4PCl_3

a) By stoichoimetry If 6 moles of Cl_2 gives 4 mole of PCl_3 , then 22 moles of Cl_2 will give: \frac{4}{6}\times 22=\frac{44}{3}=14.66 moles of PCl_3

b) If 6 moles of Cl_2 reacts with one mole of P_4 , then 22 moles of Cl_2 will react : \frac{1}{6}\times 22=\frac{11}{3}=3.66 moles of P_4

Moles of P_4  left after the reaction = 5-3.66=\frac{4}{3}=1.34 moles.

c) 0 Moles of Cl_2 , since it is present in less amount it will get completely consumed in the reaction. Hence, it is limiting reagent.



7 0
3 years ago
A field worker is exposed to a xylene for a duration of 8 weeks at 40 hrs/wk. The concentration of xylene in the workplace is 40
Andrej [43]

Answer:

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

Explanation:

Number of hours worker exposed to xylene = 40 hr/week\times 8 week = 320 hours

The concentration of xylene in the workplace =40 \mu g/m^3

The worker is inhaling air at a rate of 0.9 m^3/hr.

Amount xylene inhaled by worker in an hour :

= 40\mu g/m^3\times 0.9 m^3/hr=36 \mu g/hr

Amount xylene inhaled by worker in 320 hours:

36 \mu g/hr\times 320 hr=11,520 \mu g=11,520\times 0.001 mg=11.520 mg

1 μg = 0.001 mg

Amount xylene inhaled by worker in 320 hours = 11.520 mg

1 day = 24 hours

Amount xylene inhaled by worker in 1 day:

\frac{24}{320}\times 11.520 mg=0.864 mg

Assuming 70 kg body mass, the chronic daily intake of xylene :

\frac{0.864 mg/day}{70 kg}=0.01234 mg/ kg day\approx 0.012 mg/ kg day

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

5 0
3 years ago
A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent compos
Volgvan

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

percent composition element A=\frac{total mass of element A}{mass of compound} *100

In this case, the percent composition of fluorine is:

percent composition of fluorine=\frac{39.6 g}{60.3 g} *100

percent composition of fluorine= 65.67%

<u><em>The percent composition of fluorine is 65.67%</em></u>

4 0
3 years ago
CalcuLate the velocity of an electron ejected if 300.0 mm of light is applied to the surface. A wavelength of 795 nm has suffici
worty [1.4K]
The equation relating velocity and wavelength is written below:

v = λf
where λ is the wavelength in m while f is frequency in 1/s.

Let's determine first the frequency from the speed of light:
c = distance/time, where c is the speed of light equal to 3×10⁸ m/s
3×10⁸ m/s = (300 mm)(1 m/1000 mm)/ time
time = 1×10⁻⁹ seconds
Since f = 1/t,
f = 1/1×10⁻⁹ seconds = 10⁹ s⁻¹

Thus,
v = (795×10⁻⁹ m)(10⁹ s⁻¹)
v = 795 m/s
8 0
3 years ago
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