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4 years ago

What do you need to adjust to balance a chemical equation

Chemistry

2 answers:

Nataliya [291]4 years ago

8 0

You always adjust numbers and it will be before the element
so you Neva add or change a subscript

Charra [1.4K]4 years ago

8 0

Subscript
Is what you will need in order to adjust the balance of the chemical equation

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Determine the number of protons and neutrons in an atom of Uranium - 238

Mila [183]

Answer:

146

Explanation:

uranium is 92 and the mass number of the isotope is given as a 238 therefore it is not the two protons 92 electrons and

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3 years ago

Water is a ______ molecule and the bond between the h-o is a __ ______ bond.

omeli [17]

Answer: polar molocule

Explanation:

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3 years ago

1.4g of calcium oxide obtained by heating limestone was found to contain 0.4g of oxygen.another sample of 3.5g of calcium oxide

Leya [2.2K]

Answer:

Solution

1.4 gm of CaO⟶n0.4 gm of Oxygen

1.0 gm of O2⟶3.5 gm of Ca

According to the law of definite proportions

n/m=y/x

⇒1.4/0.4=3.5.

⇒3.5=3.5

Explanation:

6 0

3 years ago

For the following reaction, 5.20 grams of propane (C3H8) are allowed to react with 22.5 grams of oxygen gas. propane (C3H8) (g)

Sliva [168]

Answer:

15.58g of CO₂ is the maximum amount that can be produced

Explanation:

The propane reacts with oxygen as follows:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Where 1 mole of propane reacts with 5 moles of oxygen

To solve this question we need to find the moles of propane and oxygen to find limiting reactant using the chemical reaction:

Moles propane -Molar mass: 44.1g/mol-:

5.20g * (1mol / 44.1g) = 0.118 moles

Moles oxygen -Molar mass: 32g/mol-:

22.5g O₂ * (1mol / 32g) = 0.703 moles

For a complete reaction of 0.703 moles of oxygen are:

0.703 moles O₂ * (1mol C₃H₈ / 5mol O₂) = 0.141 moles of propane are necessaries. As there are just 0.118 moles of propane, propane is limiting reactant.

The moles of carbon dioxide that are produced are:

0.118 moles C₃H₈ * (3 moles CO₂ / 1 mol C₃H₈) = 0.354 moles CO₂

The maximum mass that can be produced is -Molar mass CO₂: 44.01g/mol-:

0.354 moles CO₂ * (44.01g / mol) =

15.58g of CO₂ is the maximum amount that can be produced

4 0

3 years ago

PLZ HELP *NO LINKS*

Radda [10]

Answer: (1). There are 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.

(2). The density of gaseous arsine is 3.45 g/L.

Explanation:

1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\1 atm \times 0.372 L = n \times 0.0821 L atm/mol K \times 273.15 K\\n = 0.0165 mol$

2). As number of moles are also equal to mass of a substance divided by its molar mass.

So, number of moles of Arsine $(AsH_{3})$ (molar mass = 77.95 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\0.0165 mol = \frac{mass}{77.95 g/mol}\\mass = 1.286 g$

Density is the mass of substance divided by its volume. Hence, density of arsine is calculated as follows.

$Density = \frac{mass}{volume}\\= \frac{1.286 g}{0.372 L}\\= 3.45 g/L$

Thus, we can conclude that 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP and the density of gaseous arsine is 3.45 g/L.

4 0

3 years ago