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2 years ago

1. A stone is found to have a volume of 34.9 cm3. It has a mass of 30.8 g. What is the density of the stone?

Chemistry

1 answer:

zavuch27 [327]2 years ago

7 0

Answer:

<h2>Density = 0.88 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

$Density = \frac{mass}{volume}$

From the question

mass = 30.8 g

volume = 34.9 cm³

Substitute the values into the above formula and solve

That's

$Density = \frac{30.8}{34 .9} \\ = 0.8825$

We have the final answer as

<h3>Density = 0.88 g/cm³</h3>

Hope this helps you

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A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83

guapka [62]

Answer:

a. 478.69 K

b. 939.43 $cm^{3}$

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

$V_{o}$ = 785 $cm^{3}$ = 0.000785 $m^{3}$

$T_{o}$ = 400K

$P_{o}$ = 125 Kpa = 125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 $m^{3}$ ⋅Pa⋅ $K^{-1}$ ⋅ $mol^{-1}$

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5 $T_{1}$ - (34980 + 35.5 $T_{o}$ )]

83.8 = (0.03 × 35.5) ( $T_{1}$ - 400K)

83.8 = 1.065 ( $T_{1}$ - 400)

78.69 = ( $T_{1}$ - 400)

$T_{1}$ = 400 + 78.69

$T_{1}$ = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

$V_{1}$ / $T_{1}$ = $V_{o}$ / $T_{o}$

$V_{1}$ = $V_{o}$ $T_{1}$ / $T_{o}$

$V_{1}$ = (785 × 478.69)/400

$V_{1}$ = 939.43 $cm^{3}$

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P( $V_{1}$ - $V_{o}$ ) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0

2 years ago

A sample of Xe gas is observed to effuse through a pourous barrier in 4.83 minutes. Under the same conditions, the same number o

Solnce55 [7]

Answer:

28.93 g/mol

Explanation:

This is an extension of Graham's Law of Effusion where $\frac{R1}{R2} = \sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$

We're only talking about molar mass and time (t) here so we'll just concentrate on $\sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$ . Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.

Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.

$\sqrt{\frac{M2}{131} } = \frac{2.29}{4.83}$

The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.

$(\sqrt{\frac{M2}{131} } )^2= (\frac{2.29}{4.83})^2$

${\frac{M2}{131} } = (0.47)^2$

${\frac{M2}{131} } = 0.22$

M2= 0.22 x 131

M2= 28.93 g/mol

8 0

2 years ago

What is the total number of valence electrons in a carbon atom in the ground state

kolbaska11 [484]

Carbon will have 4 valence electrons. It will have 2 in the p orbital and 2 in the s orbital. You can see this when you find the noble gas configuration of carbon which is [He]2s²2p² showing that carbon has 4 valence electrons.
I hope this helps. Let me know if anything is unclear.

6 0

2 years ago

Help pleasee I need help to write the formulas i’m so lost

pishuonlain [190]

Answer:

NH4

Explanation:

3 0

2 years ago

Please please please help

weqwewe [10]

Answer:

See explanation.

Explanation:

Hello!

In this case, since the main molecular reaction that is taken place in the beaker is:

$AgNO_3(aq)+BaI_2(aq)\rightarrow AgI(s)+Ba(NO_3)_2(aq)$

In such a way, we understand that one breaker contained silver nitrate and the other one barium iodide. Thus, the complete molecular equations turns out:

$2AgNO_3(aq)+BaI_2(aq)\rightarrow 2AgI(s)+Ba(NO_3)_2(aq)$

Now, for the complete ionic equation, we just ionize the aqueous species:

$2Ag^++2NO_3^-+Ba^{2+}+2I^-\rightarrow 2AgI(s)+Ba^{2+}+2(NO_3)^-$

Finally, for the net ionic equation we cancel out barium and nitrate ions as the spectator one because they are both sides on the equation:

$2Ag^++2I^-\rightarrow 2AgI(s)$

Best regards!

3 0

2 years ago