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Phantasy [73]
3 years ago
5

How much salt (NaCl) is carried by a river flowing at 30.0 m3/s and containing 50.0 mg/L of salt? Give your answer in kg/day.

Chemistry
1 answer:
ser-zykov [4K]3 years ago
7 0

Answer:

129,600kg/day

Explanation:

The river is flowing at 30.0m^{3}/s

1m^{3}/s = 1000L

Multiply by 1000 to convert  to L/s

flowrate of river = 30*1000 =30,000L/s

Convert L/s to litre per day by multiplying by 24*60*60

flowrate of river = 30,000 * 24*60*60 L/day

                     = 2,592,000,000L/day

if the river contains 50mg of salt  in 1L of solution

lets find how many mg of salt (X) is contained in 2,592,000,000L/day

X= \frac{ 2,592,000,000*50}{1}

X= 129,600,000,000 mg/day

convert this value to kg/day by multiply by 10^{-6}

X= 129,600kg/day

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Describe how iron turns into a brownish-red powder. Name the reactants and product that are involved.
IRISSAK [1]

Answer:

Gets exposed to oxygen (or water)

Explanation:

It's called oxidizing.

5 0
3 years ago
The rate constants of some reactions double with every 10 degree rise in temperature. Assume that a reaction takes place at 271
AfilCa [17]

Answer : The activation energy for the reaction is, 119.7 J

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 271 K

K_2 = rate constant at 281 K = 2K_1

Ea = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 271 K

T_2 = final temperature = 281 K

Now put all the given values in this formula, we get:

\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{271K}-\frac{1}{281K}]

Ea=119.7J

Therefore, the activation energy for the reaction is, 119.7 J

3 0
4 years ago
How many moles of gold are Cleveland to 1.204×1024 atoms
Whitepunk [10]

Answer:

10

Explanation:

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4 0
3 years ago
I need some help with a long chemistry problem. Anything is appreciated!
Mkey [24]

Answer:

The CSI is wrong.  

Explanation:

1. Find the volume of the pool

The formula for the volume of a cylinder is  V = πr²h .

D = 12 m; h = 10 m

r = D/2 = (12 m/2) = 6.0 m

V = πr²h = π × (6.0 m)² × 10 m = π × 36 m²× 10 m = 360π m³ = 1100 m³

= 1.1. × 10⁶ L

2. Calculate the moles of OH⁻

n = cV = 1.0 × 10⁻² mol·L⁻¹ × 1.3 × 10⁶ L = 11 000 mol of OH⁻                                                                                                                                                                                                                                                                                                                              

3. Calculate the moles of acetic acid needed for neutralization

HA + OH⁻ ⟶ A⁻ + H₂O

The molar ratio of is 1 mol HA:1 mol OH⁻, so you need 11 000 moL of acetic acid.

4. Calculate the actual moles of acetic acid

You have four 5 L jugs of acetic acid pH 2 .

Volume = 20 L

[H⁺] = 10⁻² mol·L⁻¹ = 0.01 mol·L⁻¹

(a) Set up an ICE table

                      HA + H₂O ⇌ A⁻  + H₃O⁺

I/mol·L⁻¹:          c                   0         0

I/mol·L⁻¹:     - 0.01             +0.01    +0.01

I/mol·L⁻¹:    c - 0.01             0.01     0.01

K_{\text{a}} = \dfrac{\text{[A]}^{-}\text{[H$_{3}$O$^{+}$]}}{\text{[HA]}}  = 1.76 \times 10^{-5}

(b) Calculate the concentration of acetic acid

\begin{array}{rcl}\dfrac{\text{[A]}^{-}\text{[H$_{3}$O$^{+}$]}}{\text{[HA]}}& = & 1.76 \times 10^{-5}\\\\\dfrac{0.01\times 0.01}{c}& = & 1.76 \times 10^{-5}\\\\1 \times 10^{-4} & = & 1.76 \times 10^{-5}c\\c & = & \dfrac{1 \times 10^{-4}}{1.76 \times 10^{-5}}\\\\ & = & \text{6 mol/L}\\\end{array}

The concentration of the acetic acid is 6 mol·L⁻¹

(c) Calculate the moles of acetic acid

n = \text{20 L} \times \dfrac{\text{6 mol}}{\text{1 L}} = \textbf{100 mol}

You have 100 mol of acetic acid.

The CSI is wrong.

You don't have enough acetic acid to neutralize the pool.

8 0
4 years ago
Hello can someone pls help me with this question! (Please explain why)
denis23 [38]

Answer:

letter A.

Explanation:

the rate of a chemical reaction can be raised by increasing the surface area of a solid reactant. this is done by cutting the substance into small pieces, or by grinding it into a powder.if the surface area of a reactant is inceased: more particles are exposed to the other reactant.

HOPE IT HELPS(◍•ᴗ•◍)❤

8 0
3 years ago
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