All categories

3 years ago

In the combined gas law, if the volume is decreased, what happens to the temperature if pressure remains constant?

1 answer:

8 0

The correct answer is A. In the combined gas law, if the volume is decreased and the pressure is constant, then the temperature decreases.

P1V1/ T1 = P2V2 / T2

Assume the volume decrease by half; V2 = V1/2

P1V1/ T1 = P2V1 /2 T2

Cancelling terms,

1/T1 = 1/2 T2

T2 = T1/2

Thus, the temperature decreased.

You might be interested in

If a gas is initially at a pressure of 9 atm, a volume of 21 liters, and a temperature of 253 K, and then the pressure is raised

alex41 [277]

Answer:

15.04 mL

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 21 L

V₂ = ?

P₁ = 9 atm

P₂ = 15 atm

T₁ = 253 K

T₂ = 302 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{9}\times {21}}{253}=\frac {{15}\times {V_2}}{302}$

Solving for V₂ , we get:

V₂ = 15.04 mL

3 0

3 years ago

Three elements are represented by the letters A, B, and C. the three elements have the same number of valence electrons. Element

AfilCa [17]

Answer:
A- beryllium
B- calcium
C- magnesium

Explanation
NOTE: all element in group 2 have 2 balance electrons

First let’s start with B- number of electrons= number of protons which is equal to the atomic number. therefore, the answer is calcium as it’s atomic number is 20

C- magnesium will have three energy levels considering it has 12 electrons (2,8,2).

A- beryllium is the lightest one in group 2 as it has the atomic mass of 9.0122.

3 0

3 years ago

What are some applications of quantities analysis?

andriy [413]

Answer:

Needless to say, we have faced a lot of challenges in the analysis and study of such a huge volume of data with the traditional data processing tools. To overcome these challenges, some big data solutions were introduced such as Hadoop. These big data tools really helped realize the applications of big data.

Explanation:

6 0

2 years ago

Which is the largest atom in group IV?

11111nata11111 [884]

Answer:

Explanation:

god loves you

3 0

2 years ago

1.20 x 10^22 molecules NaOH to gram

Trava [24]

Answer:

$\boxed {\boxed {\sf 0.797 \ g \ NaOH}}$

Explanation:

1. Convert Molecules to Moles

First, we must convert molecules to moles using Avogadro's Number: 6.022*10²³. This tells us the number of particles in 1 mole of a substance. In this case, the particles are molecules of sodium hydroxide.

$\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Multiply by the given number of molecules.

$1.20*10^{22} \ molecules \ NaOH *\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Flip the fraction so the molecules cancel out.

$1.20*10^{22} \ molecules \ NaOH *\frac {1 \ mol \ NaOH} {6.022*10^{23} \ molecules \ NaOH}}$

$1.20*10^{22} *\frac {1 \ mol \ NaOH} {6.022*10^{23}}}$

$\frac {1.20*10^{22} \ mol \ NaOH} {6.022*10^{23}}}$

$0.0199269345732 \ mol \ NaOH$

2. Convert Moles to Grams

Next, we convert moles to grams using the molar mass.

We must calculate the molar mass using the values on the Periodic Table. Look up each individual element.

Na: 22.9897693 g/mol
O: 15.999 g/mol
H: 1.008 g/mol

Since the formula has no subscripts, we can simply add the molar masses.

NaOH: 22.9897693+15.999+1.008=39.9967693 g/mol

Use this as a ratio.

$\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

Multiply by the number of moles we calculated.

$0.0199269345732 \ mol \ NaOH*\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

The moles of sodium hydroxide cancel.

$0.0199269345732 *\frac {39.9967693 \ g \ NaOH }{1}$

$0.0199269345732 *39.9967693 \ g \ NaOH$

$0.79701300498 \ g \ NaOH$

The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 0 tells us to leave the 7 in the hundredth place.

$0.797 \ g \ NaOH$

1.20*10²² molecules of sodium hydroxide is approximately 0.797 grams.

4 0

2 years ago