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2 years ago

Need help fast please I am timed and giving 47 points.

Chemistry

2 answers:

Leokris [45]2 years ago

6 0

Answer:

Explanation:

antiseptic1488 [7]2 years ago

3 0

Answer:

C - 3

Explanation:

This can be actually tested

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Komok [63]

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6 0

2 years ago

Scoring Scheme: 3-3-2-1 Part I. How many mL of 6 M NaOH stock solution should be used to prepare the 500 mL of 0.2 M NaOH? Pleas

Orlov [11]

Answer : The volume of 6M NaOH stock solution is, 16.7 mL

Explanation :

To calculate the volume of NaOH stock solution, we use the equation given by neutralization reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of NaOH stock solution.

$M_2\text{ and }V_2$ are the molarity and volume of NaOH.

We are given:

$M_1=6M\\V_1=?\\M_2=0.2M\\V_2=500mL$

Putting values in above equation, we get:

$6M\times V_1=0.2M\times 500mL\\\\V_1=16.7mL$

Thus, the volume of 6M NaOH stock solution is, 16.7 mL

4 0

3 years ago

0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(

Studentka2010 [4]

Answer : The concentration of $Fe^{3+}$ at equilibrium is 0 M.

Solution : Given,

Concentration of $Fe^{3+}$ = 0.0200 M

Concentration of $C_2O_4^{2-}$ = 1.00 M

The given equilibrium reaction is,

$Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)$

Initially conc. 0.02 1.00 0

At eqm. (0.02-x) (1.00-3x) x

The expression of $K_c$ will be,

$K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}$

$1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}$

By solving the term, we get:

$x=0.02M$

Concentration of $Fe^{3+}$ at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

Therefore, the concentration of $Fe^{3+}$ at equilibrium is 0 M.

4 0

2 years ago

What is the mass of a sample of alcohol (specific heat = 2.4 J/gC), if it requires 4780 J of heat to raise the temperature by 5.

insens350 [35]

The mass of a sample of alcohol is found to be = m = 367 g

Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.

Explanation:

The Energy of the sample given is q = 4780

We are required to find the mass of alcohol m = ?

Given that,

The specific heat given is represented by = c = 2.4 J/gC

The temperature given is ΔT = 5.43° C

The mass of sample of alcohol can be found as follows,

The formula is c = $\frac{q}{mt}$

We can drive value of m bu shifting m on the left hand side,

m = $\frac{q}{ct}$

mass of alcohol (m) = $\frac{4780}{(2.4)( 5.43)}$

m = 367 g

Therefore, The mass of the given sample of alcohol is

m = 367g

It requires 4780 J of heat to raise the temperature by 5.43 C in the process which yields a mass of 367 g of alcohol.

4 0

3 years ago

Molecules are different from atoms because they

tatyana61 [14]

D) Contain Chemical bonds.

5 0

2 years ago

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