All categories

3 years ago

Q4. The student started with 16.0 grams of NaCl, show the calculation for the percent error that

Chemistry

1 answer:

Vedmedyk [2.9K]3 years ago

5 0

Answer:

72.6% Error

Explanation:

% error = $\frac{Real-Measured}{Real} * 100$

$\frac{58.44-16}{58.44} *100= 72.6$

58.44 is the weight of NaCl

You might be interested in

Which is the largest atom in group IV?

11111nata11111 [884]

Answer:

Explanation:

god loves you

3 0

3 years ago

Question 1(Multiple Choice Worth 3 points)

NNADVOKAT [17]

<h3><u>Answer;</u></h3>

= 3.064 g/L

<h3><u>Explanation;</u></h3>

Using the equation;

PV = nRT , where P is the pressure,. V is the volume, n is the number of moles and T is the temperature and R is the gas constant, 0.08206 L. atm. mol−1.

Number of moles is 1 since one mole has a mass equivalent to the molar mass.

Therefore; We can find the volume and thus get the density.

V = nRT/P

= (1 × 0.08206 × 498 )/ 0.939

= 43.52 L

But; density = mass/volume

Therefore;

Density = 133.34 / 43.52

= 3.064 g/L

7 0

3 years ago

What volume, in milliliters, of calcium chloride 15.0 M stock solution would you use to

Kamila [148]

<u>Answer:</u> The volume of stock solution of calcium chloride required is 10 mL.

<u>Explanation:</u>

A solution consists of solute and solvent. A solute is defined as the component that is present in a smaller proportion while the solvent is defined as the component that is present in a larger proportion.

To calculate the amount of solute needed, the formula used is:

$C_1V_1=C_2V_2$ ....(1)

where,

$C_1\text{ and }V_1$ are the concentration and volume of stock solution of calcium chloride

$C_2\text{ and }V_2$ are the concentration and volume of diluted solution of calcium chloride

Given values:

$C_1=15.0M\\C_2=0.300M\\V_2=500mL$

Plugging values in equation 1:

$15.0\times V_2=0.300\times 500\\\\V_2=10mL$

Hence, the volume of stock solution of calcium chloride required is 10 mL.

7 0

3 years ago

At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.600 M . N 2 ( g )

alexandr1967 [171]

Answer:

0.78 M

Explanation:

First, we need to know which is the value of Kc of this reaction. In order to know this, we should take the innitial values of N2, O2 and NO and write the equilibrium constant expression according to the reaction. Doing this we have the following:

N2(g) + O2(g) <------> 2NO(g) Kc = ?

Writting Kc:

Kc = [NO]² / [N2] * [O2]

Replacing the given values we have then:

Kc = (0.6)² / (0.2)*(0.2)

Kc = 9

Now that we have the Kc, let's see what happens next.

We add more NO, until it's concentration is 0.9 M, this means that we are actually altering the reaction to get more reactants than product, which means that the equilibrium is being affected. If this is true, in the reaction when is re established the equilibrium, we'll see a loss in the concentration of NO and a gaining in concentrations of the reactants. This can be easily watched by doing an ICE chart:

N2(g) + O2(g) <------> 2NO(g)

I: 0.2 0.2 0.9

C: +x +x -2x

E: 0.2+x 0.2+x 0.9-2x

Replacing in the Kc expression we have:

Kc = [NO]² / [N2] * [O2]

9 = (0.9-2x)² / (0.2+x)*(0.2+x) ----> (this can be expressed as 0.2+x)²

Here, we solve for x:

9 = (0.9-2x)² / (0.2+x)²

√9 = (0.9-2x) / (0.2+x)

3(0.2+x) = 0.9-2x

0.6 + 3x = 0.9 - 2x

3x + 2x = 0.9 - 0.6

5x = 0.3

x = 0.06 M

This means that the final concentration of NO will be:

[NO] = 0.9 - (2*0.06)

[NO] = 0.78 M

8 0

3 years ago

If you wanted to neutralize 4.00g NaOH, what volume (l) of 1.0 M HCl would you need?

zhannawk [14.2K]

The balanced reaction between NaOH and HCl is as follows
NaOH + HCl ----> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
molar mass of NaOH is 40 g/mol
number of moles = $\frac{mass present }{molar mass}$
therefore number of NaOH moles - 4.00 g / 40 g/mol = 0.100 mol
since molar ratio of HCl to NaOH is 1:1
number of HCl moles required = 0.100 mol
Molarity of HCl is 1.0 M
this means that a volume of 1 L contains 1.0 mol
therefore volume containing 0.100 mol - 0.100 mol / 1.0 mol/L = 0.10 L
volume of HCl required = 100 mL

5 0

3 years ago