In a classroom, which comparison would a teacher most likely use for describing a mole?

Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Exp

weeeeeb [17]

Answer : The grams of $N_2$ consumed is, 89.6 grams.

Solution : Given,

Mass of $CaCN_2$ = 265 g

Molar mass of $CaCN_2$ = 80 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate the moles of $CaCN_2$ .

$\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles$

The given balanced reaction is,

$CaC_2+N_2\rightarrow CaCN_2+C$

from the reaction, we conclude that

As, 1 mole of $CaCN_2$ produces from 1 mole of $N_2$

So, 3.2 moles of $CaCN_2$ produces from 3.2 moles of $N_2$

Now we have to calculate the mass of $N_2$

$\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2$

$\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g$

Therefore, the grams of $N_2$ consumed is, 89.6 grams.

5 0

2 years ago

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Write the chemical symbol for the ion with 95 protons and 89 electrons.

qaws [65]

The element is Am and since you lose e- there must be a postive charge. Am+6 is the symbol

3 0

2 years ago

A sample of 0.6760 g of an unknown compound containing barium ions (ba2+) is dissolved in water and treated with an excess of na

notka56 [123]

Answer: 35.72 % of Barium ions will be present in the original unknown compound.

Explanation: The reaction of Barium ions and sodium sulfate is:

$Na_2SO_4(aq.)+Ba^{2+}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)$

Here, Sodium sulfate is present in excess, Barium ions are the limiting reagent because it limits the formation of product.

Now, 1 mole of barium sulfate is produced by 1 mole of Barium ions.

Molar mass of Barium sulfate = 233.38 g/mol

Molar mass of Barium ions = 137.327 g/mol

233.38 g/mol of barium sulfate will be produced by 137.323 g/mol of Barium ions, so

0.4105 grams of barium sulfate will be produced by = $\frac{137.327g/mol}{233.38g/mol}\times 0.4105g$ of Barium ions

Mass of barium ions = 0.2415 grams

To calculate percentage by mass, we use the formula:

$\% mass=\frac{\text{Mass of solute (in grams)}}{\text{Total mass of the solution(in grams)}}\times 100$

Mass of the solution = 0.6760 grams

Putting the value in above equation, we get

$\% \text{ mass of }Ba^{2+}\text{ ions}=\frac{0.2415g}{0.6760g}\times 100$

% mass of Barium ions = 35.72%.

8 0

2 years ago

The dipole moment (μ) of HBr (a polar covalent molecule) is 0.838D (debye), and its percent ionic character is 12.4 % . Estimate

myrzilka [38]

When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, a dipole is established. The size of a dipole is measured by its dipole moment (μμ). Dipole moment is measured in Debye units, which is equal to the distance between the charges multiplied by the charge (1 Debye equals 3.34×10−30Cm3.34×10−30Cm). The dipole moment of a molecule can be calculated by Equation 1.11.1:

μ = qr

where

μ⃗ μ→ is the dipole moment vector qiqi is the magnitude of the ithith charge, and r⃗ ir→i is the vector representing the position of ithith charge.

r = μ/q

r = [0.838D(3.34×10−30 C⋅m/ 1D)]/ (1.6×10−19 C) *0.124
 r = 1.41 x10^-10 m

7 0

2 years ago

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What do you think life on earth would be like if chemistry had not been put to practical use?

tatyana61 [14]

Answer:

if chemistry hadn't been put up to practical use, we wouldn't truly understand the reason why humans are humans like makes up humans (including other understandings of biology, etc) and we wouldn't be able to have the advancements we have today (vaccines, etc).

Explanation:

5 0

1 year ago