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LuckyWell [14K]

3 years ago

12

The elements potassium (K), Calcium (Ca), and Krypton(Kr) are all part of the same __________ on the periodic table. A) Diagonal

B) Group C) Period D) Column

1 answer:

Pavel [41]3 years ago

6 0

The answer is C, I believe.

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Ostrovityanka [42]

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This chemical equation represents the burning of methane, but the equation is incomplete. What is the missing coefficient in bot

Zolol [24]

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C=1 +H=4 +O=? = C=1 +O=2+? +H=?
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SOMEONE PLEASE HELP

Zinaida [17]

Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)

8. 12600 kg/m³ (3 s.f.)

Explanation:

6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.

1 kg= 1000g

1m³= 1 ×10⁶ cm³

Mass of iron bar

= 64.2g

= 64.2 ÷1000 kg

= 0.0642 kg

Volume of iron bar

= 8.16 cm³

= 8.16 ÷ 10⁶

$= 8.16 \times 10^{ - 6} \: kg$

$\boxed{density = \frac{mass}{volume} }$

Density of iron bar

$= \frac{0.0642}{8.16 \times 10^{ - 6} }$

= 7870 kg/m³ (3 s.f.)

7.

$\boxed{mass = density \: \times volume}$

Mass

= 1.16 ×28.8

= 33.408 g

= 33.4 g (3 s.f.)

8. Volume of brick

= 12 cm³

$= 12 \times 10^{ - 6} \: m^{3} \\ = 1.2 \times 10^{ - 5} \: m ^{3}$

Mass of brick

= 151 g

= 151 ÷ 1000 kg

= 0.151 kg

Density of brick

= mass ÷ volume

$= \frac{0.151} {1.25 \times 10^{ - 5} } \\ = 12600 \: kg/ {m}^{3}$

(3 s.f.)

6 0

3 years ago

calculate the enthalpy change for converting 1.00 mol of liquid water at 100 C to water vapor at 145 degree celsius for water Δ

Svetllana [295]

Answer:

q1 = mCpΔT

= 18.016g × 1.84J/g.K × (418.15-373.15)

= 1491.72 J

q2 = n×ΔH vap

= 1mol ×44.0kJ/mol

= 44KJ

∴ qtotal = q1+ q2

= 1.498kJ + 44.0kJ

= 45.498KJ

Explanation: The heat flow can be separated into steps.all that is being observed at a constant pressure,the heat flow is equal to the enthalpy.

8 0

3 years ago

A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a

svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = $62.364\text{ L. Torr }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol$

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g$

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0

3 years ago