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4 years ago

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What is a periodic element that is named after Einstein?

2 answers:

enyata [817]4 years ago

5 0

Einsteinium, atomic number 99

olga55 [171]4 years ago

4 0

Einsteinium is a synthetic element with symbol Es and atomic number 99. It is the seventh transuranic element, and an actinide. Einsteinium was discovered as a component of the debris of the first hydrogen bomb explosion in 1952, and named after Albert Einstein.

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After your done with the second step,you will be left with a clear colorless liquid. What should be done to the liquid to finish

mafiozo [28]

In order to retrieve the soluble salt from the solution, crystallization must be carried out.<span />

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3 years ago

IgorC [24]

Answer:

Sinkholes this is the answer

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3 years ago

Read 2 more answers

What would be the freezing point of a 1.7 mole aqueous ethylene glycol solution? The freezing point depression constant for wate

nata0808 [166]

Answer:

$-3.2^oC$

Explanation:

In order to answer this question, we need to be familiar with the law of freezing point depression. The law generally states that mixing our solvent with some particular solute would decrease the freezing point of the solvent.

This may be expressed by the following relationship:

$\Delta T_f=iK_fb$

Here:

$\Delta T_f=T_{initial}-T_{final}$ is the change in the freezing point of the solvent given its initial and final freezing point temperature values;

$i$ is the van 't Hoff factor (i = 1 for non-electrolyte solutes and i depends on the number of moles of ions released per mole of ionic salt);

$K_f$ is the freezing point depression constant for the solvent;

$b=\frac{n_{solute}}{m_{solvent}}$ is molality of the solute, defined as a ratio between the moles of solute and the mass of solvent (in kilograms).

We're assuming that you meant 1.7-molal solution, then:

$b=1.7 m$

Given ethylene glycol, an organic non-electrolyte solute:

$i=1$

The freezing point depression constant:

$K_f=1.86^oC/m$

Initial freezing point of pure water:

$T_{initial}=0.00^oC$

Rearrange the equation for the final freezing point and substitute the variables:

$T_f=T_o-iK_fb=0.00^oC-1\cdot1.86^oC/m\cdot1.7 m=-3.16^oC$

8 0

3 years ago

Which is a property of a mineral<br><br> A. Size<br> B. Density <br> C. Odor<br> D. Age

ollegr [7]

B. density is the right answer

3 0

3 years ago

What is the empirical formula of a hydrocarbon if complete combustion or 2.900 mg of the hydrocarbon produced 9.803 mg of CO2 an

kow [346]

Answer:

The empirical formula of the hydrocarbon is CH.

Explanation:

The following data were obtained from the question:

Mass of hydrocarbon = 2.9 mg

Mass of CO2 = 9.803 mg

Mass of H2O = 2.006 mg

Next, we shall determine the mass of carbon (C) and hydrogen (H) in the compound since hydrocarbon contains carbon and hydrogen only.

This is illustrated below:

Molar mass of CO2 = 12 + (16x2) = 12 + 32 = 44 g/mol

Mass of CO2 = 9.803 mg

Mass of C in the compound =?

Mass of C in the compound =

12/44 x 9.803

= 2.674 mg

Molar mass of H2O = (2x1) + 16 = 2 + 16 = 18 g/mol

Mass of H2O = 2.006 mg

Mass of H in the compound =

2/18 x 2.006

= 0.223 mg

Finally, we shall determine the empirical formula of the hydrocarbon as follow:

Carbon (C) = 2.674 mg

Hydrogen (H) = 0.223 mg

Divide by their molar mass

C = 2.674 /12 = 0.223

H = 0.223 / 1 = 0.223

Divide both side by the the smallest

C = 0.223/0.223 = 1

H = 0.223/0.223 = 1

Therefore, the empirical formula of the hydrocarbon is CH.

5 0

3 years ago