Help me how to draw this graphhh??ASAPPPP

Which of the following measurements is written with two significant figures?

diamong [38]

I think C. I'm not 100% sure, tbh.

8 0

2 years ago

Find the volume of a gas STP if it’s volume is 80.0 mL at 109 kpa and-12.5c

photoshop1234 [79]

Answer:

Approximately $8.38 \times 10^8\; \rm mL$ , which is the same as $8.38 \times 10^5 \; \rm L$ . Assumption: the behavior of this gas is ideal.

Explanation:

Initial state of the gas:

$T_\text{initial} = -12.5\; \rm ^\circ C = (-12.5 + 273.15)\; \rm K = 260.65\; \rm K$ .
$P_\text{initial} = \; \rm 10^9\; \rm kPa = 10^{12}\; \rm Pa$ .

STP state:

$T_\text{STP} = 0\; \rm ^\circ C = 273.15\; \rm K$ .
$P_\text{STP} = 10^5\; \rm Pa$ .

The initial state of this gas can be changed to STP state in two steps:

First, reduce the pressure from $10^{12}\; \rm Pa$ to $10^5\; \rm Pa$ .
Second, increase the temperature from $260.65\; \rm K$ to $273.15\; \rm K$ .

Assume that the gas acts like an ideal gas at all time. Also, assume that the number of gas particles did not change.

Assume that temperature stays the same when the pressure changes from $10^{12}\; \rm Pa$ to $10^5\; \rm Pa$ . By Boyle's Law, volume is inversely proportional to pressure when all other factors stay the same. In other words,

$\begin{aligned}V_\text{intermediate} &= V_\text{initial} \cdot \displaystyle \frac{P_{\text{initial}}}{P_{\text{STP}}} \\ &= 80.0\; \rm mL \times \frac{10^{12}\; \rm Pa}{10^5\; \rm Pa} = 8.00 \times 10^8\; \rm mL\end{aligned}$ .

After that, assume that pressure stays the same when the temperature changes from $\rm 260.65\; \rm K$ to $\rm 273.15\; \rm K$ . By Charles's Law, volume is proportional to pressure when all other factors stay the same. In other words,

$\begin{aligned}V_\text{STP} &= V_\text{intermediate} \cdot \displaystyle \frac{T_{\text{STP}}}{T_{\text{initial}}} \\ &= 8.00 \times 10^8\; \rm mL \times \frac{273.15\; \rm K}{260.65\; \rm K} \\ &\approx 8.38\times 10^8\; \rm mL = 8.38 \times 10^5\; \rm L\end{aligned}$ .

7 0

3 years ago

Lead ions can be precipitated from solution with KCl according to the following reaction:

MrRa [10]

Answer:

34.48 grams = Theoretical yield

The % yield is 85.85 %

Explanation:

Step 1: The balanced equation

Pb2+(aq) + 2KCl(aq) → PbCl2(s) + 2K+(aq)

For 1 mole Pb2+ consumed, we need 2 moles of KCl to produce 1 mole of PbCl2 and 2 moles of K+

Step 2: Given data

Mass of KCl = 28.8 grams

Mass of Pb2 = 25.6 grams

Mass of dried PbCl2 = 29.6 grams

Molar mass of Pb2+ = 207.2 g/mol

Molar mass of KCl = 74.55 g/mol

Molar mass of PbCl2 = 278.1 g/mol

Step 3: Calculate moles of Pb2+

Moles = mass / Molar mass

Moles Pb2+ = 25.6 grams / 207.2 g/mol = 0.124 moles

Step 4: Calculate moles of KCl

Moles = 28.8 grams / 74.55 g/mol = 0.386 moles

Step 5: Calculate limiting reactant

For 1 mole Pb2+ consumed, we need 2 moles of KCl

Pb2+ is the limiting reactant. It will completely ( 0.124 moles) be consumed. There will remain 0 moles

There will react 2* 0.124 = 0.248 moles of KCl. There will remain 0.386 - 0.248 = 0.138 moles of KCl

Step 6: Calculate number of moles of PbCl2

For 1 mole Pb2+ consumed, we need 2 moles of KCl to produce 1 mole of PbCl2 and 2 moles of K+

For 0.124 moles of Pb2+ consumed ,there will be produced 0.124 moles of PbCl2

Step 7: Calculate mass of PbCl2

Mass = Numbers of moles * Molar mass

mass PbCl2 =0.124 moles * 278.1 g/mol = 34.48 grams = Theoretical yield

Step 8: Calculate % yield

% yield = (actual yield / theoretical yield) * 100 = (29.6 grams / 34.48 grams) *100 % = 85.85 %

The % yield is 85.85 %

8 0

3 years ago

Which of these is a property of acid?

Taya2010 [7]

C. They Typically Cantain AN -Oh group

3 0

3 years ago

Who was Dmitri Mendeleev? And why was he so important?

WARRIOR [948]

Answer:

founder of the periodic law.

Subjects Of Study: periodic law

Notable Works: “The Principles of Chemistry”

Died: February 2, 1907 (aged 72) St. Petersburg ...

Born: February 8, 1834 Tobolsk Russia

Explanation:

for u bbg, an goo//

gle helps

7 0

2 years ago