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3 years ago

This element is less reactive than lithium and magnesium but more reactive than zinc. Which element is this?

Chemistry

2 answers:

Leni [432]3 years ago

8 0

I searched online since i felt bad it is B

Tamiku [17]3 years ago

7 0

The answer should be:

B. Al (Aluminium)

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Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500=193000C$ is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

$14\times 1000g=14000g$ of copper is deposited by = $\frac{193000}{63.5}\times 14000=42551181 C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

$40.0=\frac{42551181 C}{t}\\\\t=1063779sec$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $1063779s\times \frac{1hr}{3600s}=295hr$

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0

3 years ago

If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

$n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol$

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0

3 years ago

Which statement about a chemical reaction is true?

Afina-wow [57]

yes yes yes yes yes yes yes yes yes yes yes yes yes yes

7 0

3 years ago

When sulfur burns, it forms sulfur dioxide (SO2). Its chemical reaction is S + O2 → SO2.

Elenna [48]

Answer:

The mass of SO2 will be equal to the sum of the mass of S and O2.

Explanation:

This can be explained by the <em>Law of Conservation of Mass</em>. This law states that mass can neither be created nor destroyed. Knowing this, we can say that the reactants of a chemical reaction must be equal to the products.

In this case, the reactants Sulfur (S) and Oxygen (O2) must equal the mass of the product Sulfur Dioxide (SO2). Therefore, the statement <em>"The mass of SO2 will be equal to the sum of the mass of S and O2" </em>is correct.

4 0

3 years ago

Below are shown, for five metals, reduction reactions and standard electrode potential values. which of these metals is the leas

just olya [345]

Above question is incomplete. Complete question is attached below
........................................................................................................................
Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.

In present case, reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is least reactive.

On other hand, reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is most reactive.

7 0

4 years ago