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3 years ago

204,900,000,000 How many significant digits are found?

Chemistry

1 answer:

Advocard [28]3 years ago

5 0

Answer:

Explanation:

There are 4 significant figures because when there is not a decimal, then only numerical values over 0 are counted. So the numbers counted in sig figs are 2049, and none of the other 0s.

So if it was 204,900,000,000.0, then it would have 12 significant figures.

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The temperature of 950 mL of SF6 is changed from 607 K to 525 K. During this, the pressure changes from 66.95 kPa to 16.7 kPa. W

wel

Answer: 250 cause you g

Explanation:

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3 years ago

Read 2 more answers

Using the solubility of Ca(IO3)2 that you determined in pure water, calculate the value of Ksp using only concentrations, that i

Aleks04 [339]

Answer:

4.6305 * 10^-6 mol^3.L^-3

Explanation:

Firstly, we write the value for the solubility of Ca(IO3)2 in pure water. This equals 0.0105mol/L.

We proceed to write the dissociation reaction equation for Ca(IO3)2

Ca(IO3)2(s) <——->Ca2+(aq) + 2IO3-(aq)

We set up an ICE table to calculate the Ksp. ICE stands for initial, change and equilibrium. Let the concentration of the Ca(IO3)2 be x. We write the values for the ICE table as follows:

Ca2+(aq). 2IO3-(aq)

I. 0. 0.

C. +x. +2x

E. x. 2x

The solubility product Ksp = [Ca2+][IO3-]^2

Ksp = x * (2x)^2

Ksp = 4x^3

Recall, the solubility value for Ca(IO3)2 in pure water is 0.0105mol/L

We substitute this value for x

Ksp = 4(0.0105)^2 = 4 * 0.000001157625 = 4.6305 * 10^-6

5 0

3 years ago

What is the density of a substance that weighs 100g and has a volume of 68cm3?

exis [7]

D= mass/volume so it's 100/68 which equals 1.47cm3

4 0

3 years ago

What is the oxidation number of neon

vredina [299]

Answer:

Electron Configuration and Oxidation States of Neon. Electron configuration of Neon is [He] 2s2 2p6. Possible oxidation states are 0. so the answer is 0.

Explanation:

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4 years ago

typical room is 4.0 m long, 5.0 m wide, and 2.5 m high. What is the total mass of the oxygen in the room assuming that the gas i

andrey2020 [161]

Answer:

mass of oxygen gas in Kg = 15.0Kg

Explanation:

Volume of air in the room = 4.0m*5.0m*2.5m = 50m³

volume of oxygen in the room = 21/100 * 50m³ = 10.5m³

using the ideal gas equation; PV=nRT

number of moles of oxygen gas, n = PV/RT

At STP, P = 1atm, V = 10.5m³ = (10.5*1000)dm³ = 10500dm³, R = 0.082 atmdm³K⁻¹mol⁻¹, T = 273K

n = 1 * 10500/ (273 *0.082)

n = 469.04 moles

mass of oxygen gas in Kg = (no of moles * molar mass)/1000

molar mass of oxygen gas = 32g

mass of oxygen gas in Kg = (469.04 * 32)/1000

mass of oxygen gas in Kg = 15.0Kg

8 0

3 years ago