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Firlakuza [10]
3 years ago
13

The limiting reactant in a chemical reaction is the reactant __________ Select one: A. for which you have the lowest mass in gra

ms. B. which has the lowest coefficient in the balanced equation. C. which has the lowest molar mass. D. which is left over after the reaction has gone to completion. E. None of the above.
Chemistry
1 answer:
Maru [420]3 years ago
3 0

Answer:

i think its A

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Describe the process shown occurring at B, and explain what results from this.​
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6 0
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If 8.87 grams of phosphorus react with 11.43 grams of oxygen, what is the empirical fr pound formed?
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Answer: Phosphorus 3143.58=1.4 1

Oxygen 1656.42=3.5 2.5

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3 years ago
Read 2 more answers
g If 0.600 g of a gas occupies 211 mL at 24 oC and 660 mmHg of pressure, what is the molar mass of the gas
Alisiya [41]

Answer:

79.89 g/mol

Explanation:

The following data were obtained from the question:

Mass (m) of gas = 0.6 g

Volume (V) = 211 mL

Temperature (T) = 24 °C

Pressure (P) = 660 mmHg

Molar mass of gas =.?

Next, we convert 24 °C to Kelvin temperature. This can be obtained as follow:

Temperature (K) = temperature (°C) + 273

Temperature (°C ) = 24 °C

Temperature (K) = 24 + 273 = 297 K

Next, we shall convert 211 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

211 mL = 211 mL × 1 L / 1000 mL

211 mL = 0.211 L

Next, we shall convert 660 mmHg to atm. This can be obtained as follow:

760 mmHg = 1 atm

Therefore,

660 mmHg = 660 mmHg × 1 atm / 760 mmHg

660 mmHg = 0.868 atm

Next, we shall determine the number of mole of the gas. This can be obtained as shown below:

Volume (V) = 0.211 L

Temperature (T) = 297 K

Pressure (P) = 0.868 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?

PV = nRT

0.868 × 0.211 = n × 0.0821 × 297

Divide both side by 0.0821 × 297

n = (0.868 × 0.211) / (0.0821 × 297)

n = 7.51×10¯³ mole.

Finally, we shall determine the molar mass of the gas as shown below:

Mass of gas = 0.6 g

Number of mole of gas = 7.51×10¯³ mole.

Molar mass of gas =.?

Mole = mass /Molar mass

7.51×10¯³ = 0.6 / Molar mass

Cross multiply

7.51×10¯³ × molar mass = 0.6

Divide both side by 7.51×10¯³

Molar mass = 0.6 / 7.51×10¯³

Molar mass of gas = 79.89 g/mol

3 0
3 years ago
2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)
yan [13]

5.732 grams of AgCl is formed when 0.200 L of 0.200 M AGNO3 reacts with an excess of CaCl2.

Explanation:

The balanced equation:

2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)

data given:

volume of AgNO3 = 0.2 L

molarity of AgNO3 = 0.200 M

atomic weight of AgCl= 143.32 gram/mole

from the formula, number of moles can be calculated

Molarity = \frac{number of moles}{volume in litres}

number of moles of AgNO3 = 0.04

From the reaction:

2 moles of AgNO3 reacts to form 2 moles of AgCl

0.04 moles of AgNO3 reacts to form x mole of AgCl

\frac{2}{2} = \frac{x}{0.04}

= 0.04 moles of AgCl is formed

mass of AgCl formed is calculated by multiplying number of moles with atomic mass of AgCl

mass of AgCl = 0.04 x 143.32

                       = 5.732 grams of AgCl is formed.

4 0
3 years ago
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