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lyudmila [28]
3 years ago
13

Is radium flammable?

Chemistry
1 answer:
rewona [7]3 years ago
4 0

Answer:

no

Explanation:

Radium is silvery, lustrous, soft, intensely radioactive. It readily oxidizes on exposure to air, turning from almost pure white to black. Radium is luminescent, corrodes in water to form radium hydroxide. Although is the heaviest member of the alkaline-earth group it is the most volatile.

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How did Bohr describe the arrangement of particles within an atom?
alexandr1967 [171]

Answer: A

Explanation: Protons and neutrons form the nucleus of the atom, with electrons orbiting it.

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2 years ago
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Which ion in the ground state has the same electron configuration as an atom of
Sergeu [11.5K]

Answer:

Ca^+2

Explanation:

Hence, the correct option is A.

4 0
2 years ago
Acenapthalene has the empirical formula C6H5. A solution of 0.515 g of acenapthalene in 15.0 g CHCl3 boils at 62.5oC. The normal
german

Answer:

The molecular formula of an ascenapthalene is C_{12}H_{10}

Explanation:

\Delta T_b=K_b\times m

\Delta T_b=K_b\times \frac{\text{Mass of acenapthalene}}{\text{Molar mass of acenapthalene}\times \text{Mass of chloroform in Kg}}

where,

\Delta T_f =Elevation in boiling point = (62.5-61.7)^oC=0.8^oC

Mass of acenapthalene = 0.515 g

Mass of CHCl_3 = 15.0 g = 0.015 kg (1 kg = 1000 g)

K_b = boiling point constant = 3.63 °C/m

m = molality

Now put all the given values in this formula, we get

0.8^0C=3.67 ^oC/m\times \frac{0.515}{\text{Molar mass of acenapthalene}\times 0.015kg}

\text{Molar mass of acenapthalene}=155.7875 g/mol

Let the molecule formula of the Acenapthalene be C_{6n]H_{5n}

6n\times 12 g/mol+5n\times 1 g/mol=155.7875 g/mol

n = 2.0

The molecular formula of an ascenapthalene is C_{12}H_{10}

4 0
3 years ago
What is the total number of moles of oxygen atoms present in one mole of Mg(CIO3)2?
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4) 6!!!!!!!!!!!!!!
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8 0
2 years ago
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Based on the ideal gas law, there is a simple equivalency that exists between the amount of gas and the volume it occupies. at s
nordsb [41]
<span>2.10 grams. The balanced equation for the reaction is CO + 2H2 ==> CH3OH The key thing to take from this equation is that it takes 2 hydrogen molecules per carbon monoxide molecule for this reaction. And since we've been given an equal number of molecules for each reactant, the limiting reactant will be hydrogen. We can effectively claim that we have 5.86/2 = 2.93 l of hydrogen and an excess of CO to consume all of the hydrogen. So the number of moles of hydrogen gas we have is: 2.93 l / 22.4 l/mol = 0.130803571 mol And since it takes 2 moles of hydrogen gas to make 1 mole of methanol, divide by 2, getting. 0.130803571 mol / 2 = 0.065401786 mol Now we just need to multiply the number of moles of methanol by its molar mass. First lookup the atomic weights involved. Atomic weight carbon = 12.0107 g/mol Atomic weight hydrogen = 1.00794 g/mol Atomic weight oxygen = 15.999 g/mol Molar mass CH3OH = 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 g/mol So the mass produced is 32.04146 g/mol * 0.065401786 mol = 2.095568701 g And of course, properly round the answer to 3 significant digits, giving 2.10 grams.</span>
4 0
3 years ago
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