What would the mass of 7 red apples be in kilograms?

The question in the picture, I really a correct answer, no cheap answers

Delvig [45]

Answer:

94.325 g

Explanation:

We'll begin by converting 350 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

350 mL = 350 mL × 1 L /1000 mL

350 mL = 0.35 L

Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:

Volume = 0.35 L

Molarity of KC₂H₃O₂ = 2.75 M

Mole of KC₂H₃O₂ =?

Molarity = mole /Volume

2.75 = Mole of KC₂H₃O₂ / 0.35

Cross multiply

Mole of KC₂H₃O₂ = 2.75 × 0.35

Mole of KC₂H₃O₂ = 0.9625 mole

Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:

Mole of KC₂H₃O₂ = 0.9625 mole

Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)

= 39 + 24 + 3 + 32

= 98 g/mol

Mass of KC₂H₃O₂ =?

Mass = mole × molar mass

Mass of KC₂H₃O₂ = 0.9625 × 98

Mass of KC₂H₃O₂ = 94.325 g

Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g

6 0

3 years ago

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

3 0

4 years ago

Determine the mass of 2.75 moles of CaSO4. Record your work and your answer.

maw [93]

Explanation:

$RFM \: = 160 \: g \\ 1 \: mole \: weighs \: 160 \: g \\ 2.75 \: moles \: weighs \: \frac{2.75 \times 160}{1} g \\ = 440 \: g$

6 0

3 years ago

A buffer solution is composed of 7.50 mol of acid and 4.75 mol of the conjugate base. If the pKa of the acid is 4.90 , what is t

Mamont248 [21]

Answer:

The correct answer is: pH= 4.70

Explanation:

We use the <em>Henderson-Hasselbach equation</em> in order to calculate the pH of a buffer solution:

$pH= pKa + log \frac{ [conjugate base]}{[acid]}$

Given:

pKa= 4.90

[conjugate base]= 4.75 mol

[acid]= 7.50 mol

We calculate pH as follows:

pH = 4.90 + log (4.75 mol/7.50 mol) = 4.90 + (-0.20) = 4.70

7 0

3 years ago

knowing that the reaction of potassium in water is extremely exothermic, what would be the problem with adding a large piece of

anygoal [31]

Answer:

Explanation:

If the reaction is really exothermic (and it is) then the water would spatter all over the place. It would boil off if the container could hold it. It would also react according to the following reaction.

You are talking about a reaction like

2K + 2HOH = 2KOH + H2

8 0

3 years ago