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4 years ago

How many grams of oxygen are needed to react with 4.6 grams of titanium (IV) chloride

Chemistry

1 answer:

kvv77 [185]4 years ago

6 0

Answer:

1.24g

Explanation:

Given parameters:

Mass of titanium (IV) chloride = 4.6g

Unknown:

Mass of oxygen = ?

Solution:

To solve this problem, we need to write the balance reaction equation first;

TiCl₂ + O₂ → TiO₂ + Cl₂

Now to find the mass of the oxygen gas,

Express the given mass of the known TiCl₂ in moles:

Number of moles of TiCl₂ = $\frac{mass}{molar mass}$

Molar mass of TiC₂ = 47.9 + 2(35.5) = 118.9g/mol

Number of moles of TiCl₂ = $\frac{4.6}{118.9}$ = 0.04mole

From the reaction equation;

1 mole of TiCl₂ reacted with 1 mole of oxygen gas

0.04 mole of TiCl₂ will react with 0.04 mole of oxygen gas

Mass of oxygen gas = number of moles x molar mass

molar mass of oxygen gas = 2(16) = 32g/mol

Mass of oxygen gas = 0.04 x 32 = 1.24g

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3 years ago

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What would be the charge on an ion whose neutral atom has an electron configuration of 2.8.7?

kompoz [17]

Answer:

–1

Explanation:

The electronic configuration of the neutral atom is given below:

2 8 7

From the electronic configuration of the neutral atom, we can see that the atom has 7 electrons in it's outermost shell. This implies that the element needs 1 more electron to complete it's octet configuration.

Therefore the charge on the ion of the element will be –1 indicating that the atom has received 1 electron to complete it's octet configuration.

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3 years ago

The block in this illustration is floating in water, which has a density of 1.00 g/cm3 .What is a good estimate of the density o

miss Akunina [59]

Answer : The density of the block is estimated to be, $0.30g/cm^3$ .

Explanation :

Density of the water, $\rho_w=1.0 g/cm^3$

As we know that,

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where,

g = gravitational acceleration

volume of the block = v

density of the block = $\rho_o$

Weight of the Block = Buoyant force exerted by water

$\rho_o\times v\times g=\rho_w\times 30\%\text{ of volume of block}\times g$

$\rho_o\times v\times g=\rho_w\times 0.30v\times g$

$\rho_o\times v\times 9.8 m/s^2=1 g/cm^3\times 0.30v\times 9.8 m/s^2$

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3 years ago

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Pani-rosa [81]

Answer:

Option B: cooling as the lava runs down the mountain

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3 years ago

Having difficulty understanding formulas involved. Determine the quantity (g) of pure CaCl2 in 7.5 g of CaCl2•9H2O.

Luda [366]

To determine the quantity of pure CaCl2 from the given amount of CaCl2.9H2O, we use the ratio of CaCl2 to CaCl2.9H2O. From the formula the ratio would 1 is to 1 We calculate as follows:

7.5 g CaCl2.9H2O (1 mol / 273.1215 g) ( 1 mol CaCl2 / 1 mol CaCl2.9H2O ) ( <span>110.98 g / 1 mol ) = 3.05 g CaCl2

Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>

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3 years ago