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3 years ago

I need help with chemistry :(

Chemistry

2 answers:

Luda [366]3 years ago

8 0

I don’t see the problem

Yanka [14]3 years ago

4 0

Answer:

How may we help kind sir

Explanation:

and if this was for points thanks

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Fluoride ion is poisonous in relatively low amounts: 0.2 g of F- per 70 kg of body weight can cause death. Nevertheless, in orde

svetoff [14.1K]

Answer:

$\boxed{\text{(a) 14 L; (b) 711 kg}}$

Explanation:

(a) Litres of water

$\text{Mass of F}^{-} = 8.5 \times 10^{7}\text{ gal} \times \dfrac{\text{0.2 mg F}^{-}}{\text{1 kg}} = \textbf{14 mg F}^{-}\\\\\text{Volume of water} = \text{14 mg F}^{-} \times \dfrac{\text{1 L water}}{\text{1 mg F}^{-}} = \textbf{14 L water}\\\\\text{The person would have to consume $\boxed{\textbf{14 L}}$ of water per day}$

(b) Mass of NaF

$\text{Volume} =8.5 \times 10^{7} \text{gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} = 3.22 \times 10^{8}\text{ L}\\\\\text{Milligrams of F}^{-} = 3.22 \times 10^{8}\text{ L} \times \dfrac{\text{ 1 mg F}^{-}}{\text{1 L}} = 3.22 \times 10^{8}\text{ mg F}^{-}\\\\\text{kilograms of F}^{-} = 3.22 \times 10^{8}\text{ mg F}^{-} \times \dfrac{\text{1 mg F}^{-}}{10^{6}\text{kg F⁻}} = \text{322 kg F}^{-}$

1 kmol of NaF (41.99 kg) contains 19.00 kg of F⁻.

$\text{Kilograms of NaF}= \text{322 kg F}^{-} \times \dfrac{\text{41.99 kg NaF}}{\text{19.00 kg F}^{-}} = \text{711 kg NaF}\\\\\text{It will take } \boxed{\textbf{711 kg}} \text{ of NaF to treat the reservoir}$

7 0

3 years ago

Read 2 more answers

What product of the krebs cycle is considered a waste product of the reaction?

Rufina [12.5K]

The answer is carbon dioxide

5 0

3 years ago

Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of $CH_3OH$ at equilibrium = 0.0406 mole

Moles of $CO$ at equilibrium = 0.170 mole

Moles of $H_2$ at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M$

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

7 0

3 years ago

A compound made of elements A and B, has a cubic unit cell. There is an A atom at each corner of the cube and an A atom at the c

Lorico [155]

Answer:

A₅B₄

Explanation:

Since we have one atom of element A at the center of each face of the unit cell, since the unit cell is a cubic cell, we have 6 faces. Since the atom on the face of the unit cell is shared with another cell, we have half of it in the unit cell is shared So, the number of atoms per face is 1/2 atom/face × 6 faces = 4 atoms on the faces of the unit cell.

Also, we have 1 atom at each corner of the cubic unit cell. Since there are 8 corner in the cubic unit cell. Also, each atom at the corner is shared with 8 unit cells, so we have 1/8 atom per corner. So, the number of atoms per unit cell is 1/8 atom/corner × 8 corners = 1 atoms at the corners of the unit cell.

So, in total we have 4 + 1 = 5 atoms of element A in the unit cell.

Also, there are 4 atoms of element B in the unit cell.

So, the ratio of atoms of element A to element B is 5 : 4.

A:B = 5:4

So, the empirical formula of the compound containing elements A and B is A₅B₄

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3 years ago

What is the work required for the separation of air (21-mol-% oxygen and 79-mol-% nitrogen) at 25°C and 1 bar in a steady-flow p

Ronch [10]

Answer:

4oo

Explanation:

4 0

3 years ago