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2 years ago

If you wanted to prepare a 15.58 L of a 6.25 M solution, what volume of a 14.73 M solution must you start with?

Chemistry

1 answer:

Fed [463]2 years ago

4 0

Answer:

6.61 L

Explanation:

M1 = 6.25

V1 = 15.58

M2 = 14.73

M1V1 = M2V2

(6.25)(15.58) = (14.73)V2

V2 = 6.61 L

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What is the rate for the second order reaction Products when [A] = 0.371 M? (k = 0.761 M's")

AlekseyPX

Given :

Concentration of product [A] = 0.371 M .

Rate constant , $R = 0.761\ M^{-1}t^{-1}$ .

To Find :

The rate for the reaction .

Solution :

We know , for second order reaction , rate is given by :

$r=k[A]^2\\\\r=0.761\times 0.371^2\ M/t\\\\r=0.10\ M/t$

Therefore , the rate for the second order reaction is 0.1 M/t .

Hence , this is the required solution .

6 0

3 years ago

How many moles are in 6.777 x 1023 atoms of CO2. Watch your significant figures. Question 2 options: 1 atoms CO2 1.125 moles CO2

Archy [21]

Answer:

1.13moles

Explanation:

Given parameters:

Number of atoms = 6.777 x 10²³ atoms

Unknown:

Number of moles = ?

Solution:

A mole of a substance contains the avogadro's number of particles

6.02 x 10²³ particles = 1 mole

6.777 x 10²³ atoms will contain $\frac{ 6.777 x 10^{23} }{6.02 x 10^{23} }$ = 1.13moles

6 0

2 years ago

One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c

Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

$n(\mathrm{O_2}) = n(\mathrm{H}_2)$ .

Note that the mass of a gas $m$ is different from the number of gas particles $n$ in it. In particular, if all particles in this gas have a molar mass of $M$ , then:

$m = n \cdot M$ .

In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
$M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}$ .

Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
Right-hand side: $\displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9$ .

Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

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3 years ago

When the pressure that a gas exerts on a sealed container changes from 22.5 psi to ? psi, the temperature changes from 110 degre

natta225 [31]

Using Gay-Lussac's Law, pressure is proportional to (absolute) temperature in Kelvin. We first convert the temperature values to Kelvin: 110 C = 383.15 K, while 65 C = 338.15 K.
P1/T1 = P2/T2
22.5/383.15 = P2/338.15
P2 = 19.9 psi

8 0

3 years ago

What is the Bronsted acid of H2PO4- + OH-. ---- HPO42- + H2O?

RoseWind [281]

Answer:

The bronsted- Lowry acid is H₂PO₄⁻

Explanation:

Bronsted-Lowry acid donates a proton (H⁺)

H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O

In the reaction above, H₂PO₄⁻ is donating the proton to OH⁻ resulting in H₂O and the deprotonated species. This makes it a bronsted-Lowry acid.

7 0

3 years ago

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