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Vikentia [17]
3 years ago
10

A drinking water plant adds 500 grams of fluoride to a water tank containing 500,000 liters of drinking water. what is the conce

ntration of fluoride in the water in parts-per-million (ppm)?
Chemistry
1 answer:
nlexa [21]3 years ago
5 0
Answer is: <span>concentration of fluoride in the water in parts-per-million is 1 ppm.
</span>Parts-per-million (10⁻⁶) is<span> present at one-millionth of a </span>gram per gram of sample solution, f<span>or example mg/kg.
</span>m(fluoride) = 500 g · 1000 mg/g = 500000 mg.
m(water) = d(water) · V(water).
m(water) = 1 kg/L · 500000 L.
m(water) = 500000 kg.
arts-per-million = 500000 mg ÷ 500000 kg = 1 mg/kg = 1 ppm.
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When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coeffici
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The values of the coefficients would be 4, 5, 4, and 6 respectively.

<h3>Balancing chemical equations</h3>

The equation of the reaction can be represented by the following chemical equation:

ammonia (g) + oxygen (g) ---> nitrogen monoxide (g) + water (g)

  4NH_3(g)     +    5O_2(g)        --->           4NO(g)              +  6H_2O(g)

Thus, the coefficient of ammonia will be 4, that of oxygen will be 5, that of nitrogen monoxide will be 4, and that of water will be 6.

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5 0
2 years ago
What is the engine piston displacement in liters of an engine who’s displacement is listed as 430 in.²
sergeinik [125]

For an engine whose displacement is listed as 430 in.², the engine piston displacement in liters is mathematically given as

PD= 7.37 L  

<h3>What is the engine piston displacement in liters of an engine whose displacement is listed as 430 in.²?</h3>

Generally, the equation for the dimensional analysis method,\, we convert in to L is mathematically given as

l*(v/l)*l/v

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PD= 7.37 L  

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6 0
2 years ago
An unknown noble gas has a density of 5.84 g dm-3 at STP. Calculate its molar mass, and so identify the gas.
bagirrra123 [75]

The noble gas is Xenon and its molar mass is 131 g/mol.

<h3>What is the molar mass of the noble gas?</h3>

The molar mass of the noble gas is determined as follows;

Let molar mass of unknown gas be M, and mass of gas be m

Density of the noble gas, ρ = 5.8 g/dm³

density = m/V

At STP;

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From ideal gas equation:

PV = nRT

where n = m/M

PV = mRT/M

M = mRT/PV

M =  0.0821 * 273.15 * 5.84/1

Molar mass of the noble gas = 131 g/mol

The noble gas is Xenon which has molar mass approximately equal to 131 g/mol.

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