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2 years ago

Write the general electron configuration of a transition element(b) in Period 6.

Chemistry

1 answer:

aksik [14]2 years ago

6 0

The general electronic configuration of transition element is (n−1)d1−10ns1−2.

<h3>What are the transition element in period 6?</h3>

The period 6 transition metals are;

Lanthanum (La), Hafnium (Hf), Tantalum (Ta), Tungsten (W),

Rhenium (Re), Osmium (Os), Iridium (Ir), pPatinum (Pt),

Gold (Au), and Mercury (Hg).

As a rule, period 6 elements fill their 6s shells first, then their 4f, 5d, and 6p shells, in that order.

However, there are exceptions, such as gold.

To learn more about transition element, refer

https://brainly.in/question/27629573

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Which statement best describes the properties of sucrose?

crimeas [40]

Answer:

D. They are different from the properties of carbon, oxygen, and hydrogen.

Step-by-step explanation:

Carbon, oxygen, and hydrogen are all elements.

Sucrose is a compound containing carbon, oxygen, and hydrogen. Its properties are different from those of its elements.

For example, carbon is a black solid, while oxygen and hydrogen are colourless gases. Sucrose is a white solid.

6 0

3 years ago

Read 2 more answers

Thermal expansion occurs when particles of matter move farther apart as temperature increase. True or false ?

IceJOKER [234]

Answer:

true

Explanation:

3 0

3 years ago

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g) In a titration experiment, H2O2(aq) reacts with aqueous Mn

Tomtit [17]

Answer:

Oxygen in hydrogen peroxide oxidizes from -1 to 0.

Explanation:

Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.

The given reaction is shown below as:

$5 H_2O_2_{(aq)} + 2 MnO_4^-_{(aq)} + 6 H^+_{(aq)}\rightarrow 2 Mn^{2+}_{(aq)} + 8 H_2O_{(l)} + 5 O_2_{(g)}$

Manganese in $MnO_4^-$ has oxidation state of +7

Manganese in $Mn^{2+}$ has an oxidation state of +2

It reduces from +7 to +2

Oxygen in hydrogen peroxide has an oxidation state of -1.

Oxygen in molecular oxygen has an oxidation of 0.

Thus, oxygen in hydrogen peroxide oxidizes from -1 to 0.

8 0

3 years ago

Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 K. Upon combusting t

Masja [62]

Answer:

Molecular formula of hydrocarbon is: C₂H₆

Explanation:

The combusting of the hydrocarbon:

CxHy + O₂ → CO₂ + H₂O

Using gas law to obtain molar mass of the gas mixture (CO₂ + H₂O):

P/RT = n/V

Where:

P is pressure (1,208 atm)

R is gas constant (0,082 atmL/molK)

T is temperature (375 K)

n/V = 0,0393 mol/L

1,1128 g/L ÷ 0,0393 mol/L = 28,32 g/mol

Thus, average molecular weight is:

28,32 g/mol = 44,01 g/mol X + 18,02 g/mol Y

1 = X + Y

Where X is CO₂ molar percentage and Y is H₂O molar percentage.

Solving:

X = 0,397

Y = 0,603

39,7% H₂O

60,3% CO₂

With this proportion you can obtain ratio CO₂:H₂O thus:

60,3/39,7 = 1,52 So, 2 CO₂: 3H₂O

The moles of the hydrocarbon are:

PV/RT = n

P (0,959 atm)

V ( 1/5 of final volume)

T (298K)

n = 7,85x10⁻³ mol

The moles of CO₂ are:

0,0393 mol × 2 mol CO₂/ 5 mol total =0,01572.

Ratio of CO₂:CxHY =

0,01572 : 7,85x10⁻³ 2 CO₂: 1 CxHY

Doing:

1 CxHy + O₂ → 2 CO₂ + 3 H₂O

By mass balance:

1 C₂H₆ + 7/2 O₂ → 2 CO₂ + 3 H₂O

Thus, molecular formula of hydrocarbon is: C₂H₆

I hope it helps!

7 0

3 years ago

The equilibrium constant for the reaction

Hitman42 [59]

The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

Answer: The system which represents the equilibrium having value of $K_c=2.0$ is system (b)

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$