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3 years ago

9

This statement demonstrates that ionic bonds create substances that have ________ properties than the elements that created them

.

1 answer:

nignag [31]3 years ago

4 0

"different" should be the word in the blank

You might be interested in

How many moles is 22.4 liters of oxygen gas at standard temperature and pressure represent

makkiz [27]

Answer:

So 1 mole

Explanation:

PV = nRT

P = Pressure atm

V = Volume L

n = Moles

R = 0.08206 L·atm·mol−1·K−1.

T = Temperature K

standard temperature = 273K

standard pressure = 1 atm

22.4 liters of oxygen

Ok so we have

V = 22.4

P = 1 atm

PV = nRT

n = PV/RT

n = 22.4/(0.08206 x 273)

n = 22.4/22.40

n = 1 mole

7 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

3 years ago

What's the answer to equation Ba+Br2=

Andrej [43]

Answer:

BaBr2, Barium Bromide

Explanation:

7 0

4 years ago

If the theoretical yield of a reaction is 0.110 g and the actual yield is 0.104 g , what is the percent yield?

Westkost [7]

Theoretical Yield is an Ideal yield with 100 % conversion of reactant to product. It is in fact a paper work.

While,

Actual Yield is the yield which is obtained experimentally. It is always less than theoretical yield because it is not possible to have 100% conversion of reactants into products. Even some amount of product is lost while handling it during the process.

Percentage Yield is Calculated as,

%age Yield = Actual Yield / Theoretical Yield × 100

Data Given:
Actual Yield = 0.104 g

Theoretical Yield = 0.110 g

Putting Values,

%age Yield = 0.104 g / 0.110 g × 100

%age Yield = 94.54 %

3 0

3 years ago

The correct labels for the numbers in Figure 7-1 are (1) ____________________, (2) ____________________, and (3) _______________

Scrat [10]

Answer:

A horizon, B horizon, C horizon .

6 0

3 years ago