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2 years ago

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Chemistry

1 answer:

Anna71 [15]2 years ago

4 0

Answer:

This flexible ability is important because it allows the cell to survive in differing environments, such as when immersed in water over long periods of time.

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How many equivalents are present in 10 g of Ca2+? 1 0.5 1.5 2

Lynna [10]

0.5

Explanation:

Given parameters:

Mass of Ca²⁺ = 10g

unknown:

Equivalent weight = ?

Solution:

Equivalent weight that is the amount of electrons which a substance gains or loses per mole.

Ca²⁺ has +3 charge

It lost 2e⁻;

therefore;

In 1 mole of Ca²⁺, we have 2 equivalent weight

1 mol Ca²⁺ = 2eq. wts.

1 mol Ca x (40 g / 1 mol ) x (1 mol / 2 eq.wts.) = 20.0 g = 1 eq.wt.

Therefore;

10.0 g Ca²⁺ x (1 eq.wt. / 20.0 g) = 0.5 eq.wts.

learn more:

Molar mass brainly.com/question/2861244

#learnwithBrainly

8 0

2 years ago

Conner is testing the pH of a swimming pool and he finds it to be 6.5. The optimal pH for swimming is between 7.7 and 7.6. Conne

miv72 [106K]

Answer:

B.) He added a base to raise the PH

Explanation:

Took test

6 0

1 year ago

The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO

Irina-Kira [14]

Answer:

Amount of excess Carbon (ii) oxide left over = 23.75 g

Explanation:

Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

Molar mass of Fe₂O₃ = 160 g/mol;

Molar mass of Carbon (ii) oxide = 28 g/mol

From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide

450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide

Therefore the excess reactant is carbon (ii) oxide.

Amount of excess Carbon (ii) oxide left over = 260 - 236.25

Amount of excess Carbon (ii) oxide left over = 23.75 g

5 0

2 years ago

A compound has a molar mass of 90. grams per mole and the empirical formula CH2O. What is the molecular formula of this compound

k0ka [10]

Answer:

C₃H₆O₃

Explanation:

Data:

EF = CH₂O

MM = 90. g/mol

Calculations:

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{90. u}}{\text{30.03 u}} = 3.00 \approx 3$

MF = (CH₂O)₃ = C₃H₆O₃

7 0

3 years ago

The decomposition of HBr(g) into elemental species is found to have a rate constant of 4.2 ×10−3atm s−1. If 2.00 atm of HBr are

Dennis_Churaev [7]

Answer:

7,94 minutes

Explanation:

If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the reaction rate does not depend on the concentration of the reactants.

For the zero-order reactions, concentration-time equation can be written as follows:

[A] = - Kt + [Ao]

where:

[A]: concentration of the reactant A at the t time,
[A]o: initial concentration of the reactant A,
K: rate constant,
t: elapsed time of the reaction

To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.

Data:

K = 4.2 ×10−3atm/s,

[A]o=[HBr]o= 2 atm,

[A]=[HBr]=0 atm (all HBr(g) is gone)

We clear the incognita :

[A] = - Kt + [Ao]............. Kt = [Ao] - [A]

t = ([Ao] - [A])/K

We replace the numerical values:

t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes

So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).

6 0

2 years ago