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2 years ago

As you go down group one of the periodic table, the reactions become and more

Chemistry

1 answer:

Lady bird [3.3K]2 years ago

3 0

Answer:

vigorous

Explanation:

As you go down group one of the periodic table, the reactions become and more vigorous.

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What are the products of photosynthesis? A: C6H12O6 + 6H2O + 6O2 B: C6H12O6 + 3O2 + 6H2O C: C6H12O6 + 3O2 + 3H2O D: C3H6O3 + 6O2

erik [133]

A: C₆H₁₂O₆ + 6H₂O + 6O₂

6CO₂ + 12H₂O = C₆H₁₂O₆ + 6H₂O + 6O₂

8 0

3 years ago

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Water comes from different a.levels b.phases c.sources d.stages

bezimeni [28]

Answer:

C. sources

Explanation:

6 0

2 years ago

If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was

Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

$n_{acid}=n_{base}=n_{salt}$

Whereas the moles of the salt are computed as shown below:

$n_{salt}=0.021L*0.68mol/L=0.01428mol$

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

$[salt]=0.01428mol/0.0276L=0.517M$

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

$C_6H_5NH_3^++H_2O\rightleftharpoons C_6H_5NH_2+H_3O^+$

Whose equilibrium expression is:

$Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}$

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

$2.326x10^{-5}=\frac{x^2}{0.517M}$

Whereas x is:

$x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3$

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

$pH=-log(3.47x10^{-3})\\\\pH=2.46$

Regards!

6 0

2 years ago

This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press

Solnce55 [7]

Answer:

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A ( phosphine in this case ) after a time, t

[A]₀ is the initial concentration of A

k is the rate constant, and

t is the time

We also know that for a first order reaction

k = 0.693/ t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t/= 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 / 35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀ = - kt

thus:

(pPH₃ )t = 2.29 atm x e^(- 0.0198 s⁻¹ x 70.5 s ) = 0.57 atm

3 0

3 years ago

Boron is to the immediate left of carbon in the same row in the periodic table. Which of the following statements compares the r

natita [175]

Answer:

Boron has a larger radius and the protons in carbon exert more pull.

Explanation:

Remember than elements have greater radius as they are closer to the bottom left corner, so boron would have the larger radius here. Carbon has a smaller radius, which makes it easier for the protons in carbon to exert more pull.

7 0

3 years ago

Read 2 more answers