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3 years ago

Determine the molar mass of the following elements. Include units in your answer.

Chemistry

1 answer:

son4ous [18]3 years ago

8 0

Do the atomic mass times the number of atoms you have. That'll give you your answer. Hope it helps!

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How do the densities of the two objects compare?

Marizza181 [45]

Compare the density of the object in question to the density of water. If its density is less than water, it will float. For example, oak floats because its density is 0.7 g/cm³. If the density of an object is greater than water, it will sink.

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2 years ago

Elements that have atoms with stable valence electron configurations in the ground state are found in

dsp73

Answer:

Group 18, also known as the Noble Gasses

Explanation:

Atoms strive for full stability by gaining or losing electrons to get 8 valence electrons in their valence shell, but Group 18 already has 8 electrons in their valence shell, and are therefore already stable in their ground state.

5 0

2 years ago

Identify the spectator ions in this reaction. Check all that apply.

balandron [24]

CN, LI are the only two

4 0

3 years ago

Read 2 more answers

Help please I’m stuck on thsi question

vodomira [7]

Since it’s asking for chemical to thermal, B.

7 0

3 years ago

Read 2 more answers

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0

2 years ago