Calculate the volume of 0.200 m cacl2 needed to produce 2.00 g of caco

AnonymousXDX can you answer this question to

IgorC [24]

Answer:

.

Explanation:

3 0

2 years ago

Describe the position of metal ,non metals and metollieds in perodic tables

siniylev [52]

Non metals and metollids in periodic tables are the same how this helps ;)

4 0

3 years ago

Pentaoxygen monochloride formula

kicyunya [14]

Answer:

Explanation:

O5Cl

6 0

3 years ago

A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r

Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

$P\cdot V = n\cdot R\cdot T$ ,

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of gas particles in the gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume $V$ of the gas. Rearrange the ideal gas equation for volume:

$V = \dfrac{n \cdot R \cdot T}{P}$ .

Both the temperature of the gas, $T$ , and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

$T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K}$ ,
$T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}$ .

The volume of the gas is proportional to its temperature if both $n$ and $P$ stay constant.

$n$ won't change unless the balloon leaks, and
consider $P$ to be constant, for calculations that include $T$ .

$V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}$ .

Now, keep the temperature at $T_1 =258.65\;\text{K}$ and change the pressure on the gas:

$P_1 = 0.995\;\text{atm}$ ,
$P_2 = 0.720\;\text{atm}$ .

The volume of the gas is proportional to the reciprocal of its absolute temperature $\dfrac{1}{T}$ if both $n$ and $T$ stays constant. In other words,

$V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}$

(3 sig. fig. as in the question.).

See if you get the same result if you hold $T$ constant, change $P$ , and then move on to change $T$ .

6 0

2 years ago

What is the mass of a piece of iron that releases 179.85 joules of heat as it cools from 51.29 degrees Celsius to 31.42 degrees

Sedaia [141]

Answer:

20.11 g.

Explanation:

What is given?

c (specific heat of iron) = 0.450 J/g °C.

Q (heat energy) = 179.85 J.

ΔT (change of temperature) = |31.42 °C - 51.29 °C| = 19.87 °C.

What do we need? Mass of iron (m)

Step-by-step solution:

Let's see the formula of specific heat:

$c=\frac{Q}{m\cdot\Delta T}$

Where c is specific heat, Q is heat energy, m is mass and ΔT is the change of temperature.

We just have to solve for 'm' to find the mass of iron and replace the given data that we have, like this:

$m=\frac{Q}{c\cdot\Delta T}=\frac{179.85\text{ J}}{0.450\text{ }\frac{J}{g\cdot\degree C}\cdot19.87\degree C}=20.11\text{ g.}$

The mass of the iron would be 20.11 g.

5 0

1 year ago