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kozerog [31]
3 years ago
14

Please help me with this honestly I'm clueless It's 7th science work......please help

Chemistry
2 answers:
olganol [36]3 years ago
6 0
Hiii
3. describe the physical characters that can be observed: phenotype

6. a specific characteristic of an organism: trait

7. a section of DNA that provides instructions for specific traits: genes

8. allele expressed in the phenotype even if only one copy is present in the genotype: dominant

hope this helped :)
denpristay [2]3 years ago
4 0

Answer:

describe the physical characters that can be observed: phenotype

a specific characteristic of an organism: trait

a section of DNA that provides instructions for specific traits: genes

allele expressed in the phenotype even if only one copy is present in the genotype: dominant

Explanation:

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If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0
3 years ago
Cacl2+na3po4 ca3po42 + nacl
Alex787 [66]
<h3>Answer:</h3>

3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl

<h3>Explanation:</h3>

We are given the Equation;

CaCl₂ + Na₃PO₄→ Ca₃(PO₄)₂ + NaCl

Assuming the question requires us to balance the equation;

  • A balanced chemical equation is one that has equal number of atoms of each element on both sides of the equation.
  • Balancing chemical equations ensures that they obey the law of conservation of mass in chemical equations.
  • According to the law of conservation of mass in chemical equation, the mass of the reactants should always be equal to the mass of the products.
  • Balancing chemical equations involves putting appropriate coefficients on the reactants and products.

In this case;

  • To balance the equation we are going to put the coefficients 3, 2, 1, and 6.
  • Therefore; the balanced equation will be;

3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl

5 0
3 years ago
The density of H2O2 is 1.407 g/mL, and the density of O2 is 1.428 g/L. How many liters of O2 can be made from 55 mL H2O2
balandron [24]

Explanation:

mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g

molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol

moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2

For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).

moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2

Now, you need the temperature.  If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L.  Without temperature you are not really able to continue.  I will assume you are at STP.

Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.

which is  53 mL.

8 0
2 years ago
4 g of ag2so4 will dissolve in 1l of water. calculate the solubility product (ksp) for silver (i) sulfate.
Bas_tet [7]
Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.

7 0
3 years ago
Two processes are described below:
UNO [17]

Answer:

i would say D i just did this but i kinda forgot so sorry if im wrong or A

Explanation:

4 0
3 years ago
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