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2 years ago

1. Calculate the molar mass of sucrose (C12 H11 O22)?

Chemistry

1 answer:

ohaa [14]2 years ago

4 0

Answer:

=342g

Explanation:

atomic mass of C = 12g

atomic mass of H = 1g

atomic mass of O = 16g

Solution;

C12 H22 O11

= 12 (12) + 22 (1) + 11(16)

= 144+ 22 + 176

= 342g

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

Be sure to answer all parts.

Llana [10]

Answer:

16 g has 2 sig figs

6701g has 4 sig figs

560 has 2 sig figs

Explanation:

4 0

3 years ago

How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t

ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

= 2.75 x 10²³ molecules XeF₆.

8 0

2 years ago

Mike mixes two chemicals in a container. The container quickly becomes very hot.

Kipish [7]

The chemicals are reactive with one another

7 0

2 years ago

Read 2 more answers

The graph given below shows the motion of four runners P, Q, R, and S in a 5 km marathon.

wel

Answer:-

As we can see from the graphical data,

The distance covered by all the four runners is the same 5 km.

Among the four athletes, Athlete P covers the distance in under three hours.

It is the minimum time taken among the four athletes.

Thus Athlete P covers the 5 km distance in the minimum amount of time.

We know that speed = $\frac{Distance}{Time}$

Since time taken for P is minimum, his speed is the maximum. P ran the fastest.

Time taken by Q = 4.5 hours.

Speed of Q = $\frac{Distance}{Time taken by Q}$

= $\frac{5km}{4.5hour}$

= 1.1 km/ hr

Time taken by R = 6 hours.

Speed of R = $\frac{Distance}{Time taken by R}$

= $\frac{5km}{6hour}$

= 0.8 km/ hr

3 0

3 years ago