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3 years ago

Significant figures

Chemistry

1 answer:

Sergeu [11.5K]3 years ago

4 0

Answer:

go look for that on google

Explanation:

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Please help me answser this, its a photo<br><br>o element<br>o subscribe<br>o coefficient

Oliga [24]

Answer:

3 atoms of (C)carbon, 5 atoms of (H)hydrogen and 2 atoms of (O)oxygen

Explanation:

i don't know what you mean by subscribe

and i don't know what a coefficient is

4 0

3 years ago

A certain metal M crystallizes in a lattice described by a body-centered cubic (bcc) unit cell. The lattice constant (a, the edg

Furkat [3]

Answer:

radius = 156 pm

Explanation:

The relation between radius and edge length of unit cell of BCC is

r=a $\sqrt{3}$ /4

Given

a = 360 pm

Therefore

r = r = radius = 360 $\sqrt{3}$ /4= 155.88 pm

156 pm

3 0

3 years ago

8. Which of the following are reasons why Index fossils are chosen for determining ages of rocks? Select all that apply .

igomit [66]

Answer:

B. Distributed over a very wide range

D. Limited life existence in geologic time.

Explanation:

Index fossils are very unique fossils that helps in determining the relative ages of rocks and biostratigraphic correlation. They are usually called guide fossils.

Fossils are the preserved remains of dead organisms found in rocks.
Index fossils are a special class of fossils with the following properties;

They are widely distributed.
They have a short stratigraphic range.
They show rapid evolution trends

4 0

3 years ago

Read 2 more answers

A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .

Rewrite as a quadratic equation and solve for $x$ :

$x\cdot(x + 5.00\times 10^{-4}) = (1.8\times 10^{-5} )\cdot (0.100 - x)$

$x\approx 0.00111$ .

The pH of a solution depends on its H⁺ concentration.

At equilibrium

$[\text{H}^{+}\;(aq)] = 5.00\times 10^{-4}\;\text{M} + x\;\text{M} = 0.00161\;\text{M}$ .

$\text{pH} = -\log{[\text{H}^{+}]} = 2.79$ .

5 0

2 years ago

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

2 years ago

Significant figures￼

Significant figures