All of the following are true about the Earth’s inner core EXCEPT

How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce

poizon [28]

W=0.360
p=1.18g/mL
pH=2.12
v=14.0 L
M(HCl)=36.46 g/mol
v₀-?

1) pH=-lg[H⁺]

[H⁺]=c(HCl)=10^(-pH)

n(HCl)=v[H⁺]=v*10^(-pH)

2) n(HCl)=m(HCl)/M(HCl)=wv₀p/M(HCl)

3) v*10^(-pH)=wv₀p/M(HCl)

v₀=v*10^(-pH)M(HCl)/(wp)

v₀=14.0*10^(2.12)*36.46/(0.360*1.18)=9.115 mL

Approximately 9.1 mL of concentrated solution required.

3 0

4 years ago

What is 105.6 mmhg= ? Torr

Nata [24]

Mm of Hg is the same as Torres, therefore, it should be 105.6 Torr

7 0

3 years ago

a 182.4g sample of germanium-66 is left undisturbed for 22.5 hours. At the end of that period, only 5.70g remain. What is the ha

Nana76 [90]

Answer:

Approximately $4.5\; \text{hours}$ .

Explanation:

Calculate the ratio between the mass of this sample after $22.5\; \text{hours}$ and the initial mass:

$\displaystyle \frac{5.70\; \rm g}{182.4\; \rm g} \approx 0.03125$ .

Let $n$ denote the number of half-lives in that $22.5\; \text{hours}$ (where $n\!$ might not necessarily be an integer.) The mass of the sample is supposed to become $(1/2)$ the previous quantity after each half-life. Therefore, if the initial mass of the sample is $1\; \rm g$ (for example,) the mass of the sample after $\! n$ half-lives would be ${(1/2)}^{n}\; \rm g$ . Regardless of the initial mass, the ratio between the mass of the sample after $n\!\!$ half-lives and the initial mass should be ${(1/2)}^{n}$ .

For this question:

${(1/2)}^{n}} = 0.03125$ .

Take the natural logarithm of both sides of this equation to solve for $n$ :

$\ln \left[{(1/2)}^{n}}\right] = \ln (0.03125)$ .

$n\, [\ln(1/2)] = \ln (0.03125)$ .

$\displaystyle n = \frac{\ln(0.03125)}{\ln(1/2)} \approx 5$ .

In other words, there are $5$ half-lives of this sample in $22.5\; \text{hours}$ . If the length of each half-life is constant, that length should be $(1/5) \times 22.5\; \text{hours} = 4.5\; \text{hours}$ .

4 0

3 years ago

C3H8 + 5O2 → 3CO2+ 4H2O, if 5.75L of oxygen are consumed in the above reaction, how many L of carbon dioxide are produced?

anzhelika [568]

Answer: 3.45 L carbon dioxide are produced

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given volume}}{\text {Molar volume}}=\frac{5.75L}{22.4L}=0.257moles$

$C_3H_8+5O_2(g)\rightarrow 3CO_2+4H_2O$

According to stoichiometry :

5 moles of $O_2$ produce = 3 moles of $CO_2$

Thus 0.257 moles of $O_2$ will produce= $\frac{3}{5}\times 0.257=0.154moles$ of $CO_2$

Volume of $CO_2=moles\times {\text {Molar volume}}=0.154moles\times 22.4L/mol=3.45$

Thus 3.45 L carbon dioxide are produced

6 0

3 years ago

An old sample of concentrated sulfuric acid to be used in the laboratory is approximately 98.1 percent h2so4 by mass. calculate

Airida [17]

Basis: 100 mL solution

From the given density, we calculate for the mass of the solution.

density = mass / volume
mass = density x volume

mass = (1.83 g/mL) x (100 mL) = 183 grams

Then, we calculate for the mass H2SO4 given the percentage.

mass of H2SO4 = (183 grams) x (0.981) = 179.523 grams

Calculate for the number of moles of H2SO4,
moles H2SO4 = (179.523 grams) / (98.079 g/mol)
moles H2SO4 = 1.83 moles

Molarity:
M = moles H2SO4 / volume solution (in L)
= 1.83 moles / (0.1L ) = 18.3 M

Molality:
m = moles of H2SO4 / kg of solvent
= 1.83 moles / (183 g)(1-0.983)(1 kg/ 1000 g) = 588.24 m

8 0

3 years ago