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hodyreva [135]
3 years ago
9

Which of the following elements has two valence electrons?

Chemistry
1 answer:
lara31 [8.8K]3 years ago
5 0
Beryllium, Magnesium, Calcium....etc have two valence electrons
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what mass of Fe2O3 is produced in the reaction in the table above mass of reactants= 223.4 g Fe+96.0 g O2. and mass of products=
Ganezh [65]

Answer:

AS ACCORDING TO THE LAW OF MASS CONSERVATION

REACTANTS =PRODUCTS

THEREFORE,

223.4+96=MASS OF FE2O3

=319.4 FE203

4 0
3 years ago
If 1.50g lead(II) nitrate is reacted with 1.75g sodium chromate what is the theoretical yield of the precipitate?
egoroff_w [7]

Answer:

1.46g of PbCrO₄ are the theoretical yield

Explanation:

Theoretical yield is defined as the maximum amount of products that could be produced (Assuming a yield of 100%).

The reaction of Lead (II) nitrate with sodium chromate is:

Pb(NO₃)₂(aq) + Na₂CrO₄(aq) → PbCrO₄(s) + 2NaNO₃ (aq)

First, we need to find molar mass of each reactant in order to determine limiting reactant (As the reaction is 1:1, the reactant with the lower number of moles is the limiting reactant). The moles of the limiting reactant = moles of Lead (II) chromate (The precipitate):

<em>Moles Pb(NO₃)₂ -Molar mass: 331.21g/mol-</em>

1.50g * (1mol / 331.21g) = 4.53x10⁻³ moles Pb(NO₃)₂

<em>Moles Na₂CrO₄ -Molar mass: 161.98g/mol-</em>

1.75g * (1mol / 161.98g) = 0.0108 moles

Pb(NO₃)₂ is limiting reactant and moles of PbCrO₄ are 4.53x10⁻³ moles. The mass is:

4.53x10⁻³ moles PbCrO₄ * (323.19g / mol) =

<h3>1.46g of PbCrO₄ are the theoretical yield</h3>
7 0
3 years ago
Is blood acid base or neutral
katen-ka-za [31]
It is usually neutral becase its is close to neutral number 7
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4 years ago
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7) Facilitated diffusion is
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The passive movement of molecules or ions across a plasma membrane by means of a transports protein located in the plasma membrane
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If the value of Kp for N2+3H2...2NH3 is 4/27, then for which of following reaction the value of Kp is reciprocal of above reacti
jek_recluse [69]

Answer:

C. 2O₃ ⇌ 3O₂

Explanation:

Kp is the equilibrium constant calculated from the partial pressures of a gas-phase reaction equation.

For a general gas-phase reaction aA + bB ⇌ nC + xD

the expression for the Kp = (pC)ⁿ(pD)ˣ / (pA)ᵃ(pB)ᵇ

where pA = partial pressure of A; pB = partial pressure of B; pC = partial pressure of C; pD = partial pressure of D

From the given reaction in equilibrium;  N₂ + 3H₂  ⇌ 2NH₃

Kp = (pNH₃)² / (pN₂)¹ * (pH₂)³ = 4/7

(pNH₃)² / (pN₂)¹ * (pH₂)³ = (2)²/ (1)¹ * (3)³

Therefore, number of mole of reactants and products is equivalent to partial pressure.

A. 2SO₂ ⇌ O₂ + 2SO₃

pSO₂ = 2, pO₂ = 1, pSO₃ = 2,

Kp =  2²/ (2² * 1²) = 4/4 = 1

B. N₂O₄ ⇌ 2NO₂

pN₂O₄ = 1, pNO₂ = 2

Kp = 2²/1² = 4

C. 2O₃ ⇌ 3O₂

pO₃ = 2, pO₂ = 3

Kp = 3³/2² = 27/4

D. PCl₅ ⇌ PCl₃ + Cl₂

pPCl₅ = 1, pPCl₃ = 1, pCl₂ = 1

Kp = (1¹ * 1¹) / 1¹ = 1

7 0
4 years ago
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