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2 years ago

15

Chemistry

1 answer:

Vinil7 [7]2 years ago

5 0

Answer:

barium and silicon has same valence electrons

Explanation:

barium-2,8,18,18,8,2

neon-2,8

silicon-2,8,2,2

carbon-2,4

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. In a separate experiment, the molar mass of nicotine is found to be somewhere between 150 and 180 g/mol. Calculate the molar m

stealth61 [152]

Answer:

Explanation:

The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:

Nicotine has the formula $C_xH_yN_z$ . To determine its composition, a sample is burned in excess oxygen, producing the following results:

1.0 mol of CO₂

0.70 mol of H₂O

0.20 mol of NO₂

Assume that all the atoms in nicotine are present as products

<h2>Solution</h2>

To find the empirical formula you need to find the moles of C, H, and N in each of the compound.

1.0 mol of CO₂ has 1.0 mol of C

0.70 mol of H₂O has 1.4 mol of H

0.20 mol of NO₂ has 0.20 mol of N

Thus, the ratio of moles is:

C: 1.0

H: 1.4

N: 0.20

Divide all by the smallest number: 0.20

C: 1.0 / 0.20 = 5

H: 1.4 / 0.20 = 7

N: 0.20 / 0.20 = 1

Hence, the empirical formula is C₅H₇N

Find the mass of 1 mole of units of the empirical formula:

C: 5mol × 12g/mol = 60g

H: 7mol × 1g/mol = 7 g

N: 1 mol × 14g/mol = 14g

Total mass = 60g + 7g + 14g = 81g

Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.

Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.

5 0

3 years ago

If there is 5 atoms of oxygen in the reactant, how many oxygen atoms must be in the product

tiny-mole [99]

Answer:

Explanation:

It requires 20 chracters to fill this answer

4 0

2 years ago

Which of the following characteristics is NOT true of cold fronts?

I am Lyosha [343]

I'm not 100% sure that this is right but I think D. is correct.

6 0

2 years ago

Read 2 more answers

The solution in the two arms of the U-tube are separated by a membrane that is permeable to water and glucose but not to sucrose

Gnom [1K]

Answer:

The liquid level will rise in Side A and drop in Side B.

Step-by-step explanation:

The membrane is impermeable to sucrose, but permeable to sucrose and water.

Side A Side B

2 M sucrose 1 M sucrose

1 M glucose 2 M glucose

(a) Ignoring osmotic effects

The glucose will diffuse spontaneously from the side with higher concentration to that of lower concentration until equilibrium is established. There is no change in volume on either side.

At this point, we have

Side A Side B

2 M sucrose 1 M sucrose

1.5 M glucose 1.5 M glucose

=====

(b) With osmotic effects

The solute concentration on Side A is greater than on Side B.

Water will diffuse into Side A.

The liquid level will rise in Side A and drop in Side B.

8 0

3 years ago

50.g of NaNO3 was dissolved in 1250 mL of water. what is the molality of the solution? [ Molar mass of NaNO3 = 85 g/mol

Viefleur [7K]

Answer:

Approximately $0.47\; \rm mol \cdot L^{-1}$ (note that $1\; \rm M = 1 \; \rm mol \cdot L^{-1}$ .)

Explanation:

The molarity of a solution gives the number of moles of solute in each unit volume of the solution. In this $\rm NaNO_3$ solution in water,

Let $n$ be the number of moles of the solute in the whole solution. Let $V$ represent the volume of that solution. The formula for the molarity $c$ of that solution is:

$\displaystyle c = \frac{n}{V}$ .

In this question, the volume of the solution is known to be $1250\; \rm mL$ . That's $1.250\; \rm L$ in standard units. What needs to be found is $n$ , the number of moles of $\rm NaNO_3$ in that solution.

The molar mass (formula mass) of a compound gives the mass of each mole of units of this compound. For example, the molar mass of $\rm NaNO_3$ is $85\; \rm g \cdot mol^{-1}$ means that the mass of one mole of

$\displaystyle n = \frac{m}{M}$ .

For this question,

$\begin{aligned}&n\left(\mathrm{NaNO_3}\right) \\ &= \frac{m\left(\mathrm{NaNO_3}\right)}{M\left(\mathrm{NaNO_3}\right)}\\&= \frac{50\; \rm g}{85\; \rm g \cdot mol^{-1}} \\& \approx 0.588235\; \rm mol\end{aligned}$ .

Calculate the molarity of this solution:

$\begin{aligned}c &= \frac{n}{V} \\&= \frac{0.588235\; \rm mol}{1.250\; \rm L} \\&\approx 0.47\;\rm mol \cdot L^{-1}\end{aligned}$ .

Note that $1\; \rm mol \cdot L^{-1}$ (one mole per liter solution) is the same as $1\; \rm M$ .

8 0

3 years ago