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3 years ago

Give me an example of oxide ore

Chemistry

2 answers:

user100 [1]3 years ago

7 0

example of oxide ore are hematite ,magnetite,chromite, zincite

Natalka [10]3 years ago

6 0

Explanation:

Several species of oxide minerals are also quite useful for industrial purposes, and hold significant economic value. These include chief ores such as Hematite and Magnetite (Iron), Chromite (Chromium), Manganite (Manganese), Cassiterite (Tin), Ilmenite and Rutile (Titanium), Franklinite and Zincite (Zinc).

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How many milliliters of 0.121 M HCl are needed to neutralize 44.3 mL 0.274 M Ba(OH)2? Write the equation and don't forget to bal

Sergeeva-Olga [200]

Answer:

100 mL is the volume of HCl needed to neutralize the 44.3 of Ba(OH)₂

Explanation:

This is the chemical equation:

2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

Formula for neutralization is:

Molarity of acid . Volume of acid = Molarity of base . Volume of base

0.121 . Volume of acid = 0.274 . 44.3

Volume of acid = (0.274 . 44.3) / 0.121 → 100

6 0

3 years ago

A reaction between substances Y and Z is

Anuta_ua [19.1K]

Answer : The value of activation energy for this reaction is 108.318 kJ/mol

Explanation :

The Arrhenius equation is written as:

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

Taking logarithm on both the sides, we get:

$\ln k=-\frac{Ea}{RT}+\ln A$ ............(1)

where,

k = rate constant = $2.95\times 10^{-3}L/mol.s$

Ea = activation energy = ?

T = temperature = 435 K

R = gas constant = 8.314 J/K.mole

A = pre-exponential factor = $3.00\times 10^{+10}L/mol.s$

Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:

$\ln (2.95\times 10^{-3}L/mol.s)=-\frac{Ea}{8.314J/K.mol\times 435K}+\ln (3.00\times 10^{10}L/mol.s)$

$Ea=108318.365J/mol=108.318kJ/mol$

Therefore, the value of activation energy for this reaction is 108.318 kJ/mol

3 0

3 years ago

Yeast converts glucose to ethanol and carbon dioxide during a process known as anaerobic fermentation. The chemical reaction is

Korolek [52]

Answer:

102g

Explanation:

To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.

In this case, the given equation is already balanced.

$\tex{C_6H_{12}O_6} \longrightarrow 2 \ C_2H_6O + 2 \ CO_2$

From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.

Mole= Mass ÷Mr

Mass= Mole ×Mr

Method 1: using the mass of glucose

Mr of glucose

= 6(12) +12(1) +6(16)

= 180

Moles of glucose reacted

= 200 ÷180

= $\frac{10}{9}$ mol