All categories

2 years ago

How many liters of CH, gas are in 5.40 moles of CH, gas at STP? (3 sig figs)

Chemistry

1 answer:

charle [14.2K]2 years ago

5 0

Answer:

Try this link https://chem.libretexts.org/Courses/Mount_Aloysius_College/CHEM_100%3A_General_Chemistry_(O'Connor)/08%3A_Solids_Liquids_and_Gases/8.E%3A_Solids_Liquids_and_Gases_(Exercises)

You might be interested in

Which of the following rules is applicable for balancing a chemical equation?

Ber [7]

Answer:

A.) Change only the coefficients

Explanation:

An equation is balanced when there is an equal quantity of each type of element on both sides of a reaction. When balancing an equation, the only way to manipulate the amounts of each element is by changing the coefficient values. The coefficients alter the amount of each molecule in the reaction.

The subscripts cannot be altered. If you were to change the subscripts, you would be altering the amount of atoms in a particular molecule.

7 0

1 year ago

A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is us

Zinaida [17]

Answer:

A. 96.3 mg/dL

Absolute error: 5.7 mg/dL

Relative error: 5.6%

B. 97.2 mg/dL

Absolute error: 4.8 mg/dL

Relative error: 4.7%

C. 104.8 mg/dL

Absolute error: 2.8 mg/dL

Relative error: 2.7%

D. 111.5 mg/dL

Absolute error: 9.5 mg/dL

Relative error: 9.3%

E. 110.5 mg/dL

Absolute error: 8.5 mg/dL

Relative error: 8.3%

Explanation:

The formula for the absolute error is:

Absolute error = |Actual Value - Measured Value|

The formula for the relative error is:

Relative error = |Absolute error/Actual value|

In your exercise, we have that

Actual Value = 102.0 mg/dL

A. 96.3 mg/dL:

$E_{ABS} = |102.0 - 96.3| = 5.7mg/dL$

$E_{R} = \frac{5.7}{102} = 0.056 = 5.6%$

B. 97.2 mg/dL

$E_{ABS} = |102.0 - 97.2| = 4.8mg/dL$

$E_{R} = \frac{4.8}{102} = 0.047 = 4.7%$

C. 104.8 mg/dL

$E_{ABS} = |102.0 - 104.8| = 2.8mg/dL$

$E_{R} = \frac{2.8}{102} = 0.027 = 2.7%$

D. 111.5 mg/dL

$E_{ABS} = |102.0 - 111.5| = 9.5mg/dL$

$E_{R} = \frac{9.5}{102} = 0.093 = 9.3%$

E. 110.5 mg/dL

$E_{ABS} = |102.0 - 111.5| = 8.5mg/dL$

$E_{R} = \frac{8.5}{102} = 0.083 = 8.3%$

4 0

3 years ago

The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of NaCl. A white precipitate of PbCl2

NNADVOKAT [17]

PbCl₂(aq) → Pb²⁺(aq) + 2Cl⁻(aq)

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

Pb²⁺(aq) + 2Cl⁻(aq) ⇄ PbCl₂(s)

At increase the concentration of chloride ions - concentration of lead ions decreases, the lead chloride is formed.

5 0

3 years ago

Read 2 more answers

What is the oxidation number of tin (Sn) in the compound Na2SnO2? A. -2 B. 0 C. +2 D. +3

Allisa [31]

O.N. of Na = +1

O.N. of O = -2

Let, O.N. of Tin = x

1*2 + x + -2*2 = 0

2+x-4 = 0

x-2 = 0

x = 2