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3 years ago

10

Calculate the mass (in g) of 2.1 x 1024 atoms of W.

1 answer:

vfiekz [6]3 years ago

6 0

Set up your units to algebraically cancel out the original unit, and end up with the units of the answer in the final numerator. We'll first convert to moles by using Avogadro's number, then the atomic mass of tungsten (W) from the periodic table, since the mass, in grams, of one mole of any element is the atomic mass number:
2.1x10∧24 atoms W x (1 mole/6.022x10∧23 atoms) x (183.84g/1 mole) = 641.089 grams
Since your given number has only two significant figures (2.1), your answer should also only express in 2 sig figs: 6.4x10∧2 grams

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8CO + 17H2 → C8H18 + 8H2O

nekit [7.7K]

$\frac{6.50mole CO}{1} \frac{17mole H2}{8mole CO}\frac{2 gram H2}{1mole H2} =27.625$

3 sig figs= 27.6

4 0

3 years ago

if a solution measured with the miscalibrated meter (above) gave a reading of 6.50 at 20oC, what should be the true pH (give 2 d

Andreas93 [3]

Answer:

Explanation:

Ph is the major of acidity or basicity of a solution. On a scale of 0 to 14

PH = -log 10 ^(H+)

Therefore PH= -log 10^(6.5) =0.81

Therefore PH of the solution is 0.81

This solution is therefore an acidic solution

7 0

3 years ago

determine the average volume from the following volume reading: 3.00ml, 2.0ml, 2.987 x 10^-3 l, and 3.4856 ml

lesya [120]

Answer:

The average of given values is 2.1221 ml

Explanation:

Given data:

Given measurements = 3.00 ml , 2.0 ml, 2.987 × 10⁻³ml , 3.4856 ml

Average value of given measurements = ?

Solution:

Formula:

Average value = sum of all measurement / total number of measurements

2.987 × 10⁻³ml = 0.002987 ml

Now we will put the values.

Average value = 3.00 ml + 2.0 ml + 0.002987 ml+ 3.4856 ml / 4

Average value = 8.488587 ml / 4

Average value = 2.1221 ml

The average of given values is 2.1221 ml

8 0

3 years ago

Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benz

stiv31 [10]

Answer: Freezing point of a solution will be $-1.16^0C$

Explanation:

Depression in freezing point is given by:

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

$\Delta T_f=T_f^0-T_f=(5.50-T_f)^0C$ = Depression in freezing point

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

$K_f$ = freezing point constant = $5.12^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

$(5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}$

$(5.50-T_f)^0C=6.68$

$T_f=-1.16^0C$

Thus the freezing point of a solution will be $-1.16^0C$

3 0

3 years ago

How are elements arranged in the Periodic Table? A. by their properties B. alphabetically C. by how much they are worth D. in th

algol13

Im on the same test.

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3 years ago