2 years ago

Which of the responses contains all the true

Chemistry

1 answer:

jekas [21]2 years ago

7 0

What responses am i supposed to chose from?

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How many half lives would pass if a mass of copper-66 decayed for 51 minutes? The half life of Cu-66 is 5.10 minutes.

Sveta_85 [38]

Divide 51 by 5.1, it's 10 half lives

4 0

3 years ago

Identify both the cellular structure that assembles these proteins and the kinds of molecules that are used as the building bloc

Blizzard [7]

The basic building blocks of proteins are the amino acids. There are 20 amino acids in a protein that we consume and the protein in our body. These amino acids link together to form large molecules. The 20 amino acids are divided into two groups called the essential and non-essential amino acids.

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3 years ago

How many mole ratios can be correctly obtained from the chemical equation 2Al2O3(l) ® 4Al(s) + 3O2(g)?

scoundrel [369]

Answer:

Explanation:

3 0

3 years ago

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

5 0

3 years ago

What is the expected mass (in kg) of a 10 over 5 B isotope? 1 proton = 1.6726 × 10-27 kg 1 neutron = 1.6749 × 10-27 kg A.1.67 ×

Ira Lisetskai [31]

$^{10}_5 B$
An atom of this isotope contains 5 protons and 10-5=5 neutrons.

$5 \times 1.6726 \times 10^{-27} + 5 \times 1.6749 \times 10^{-27} = \\
8.363 \times 10^{-27} + 8.3745 \times 10^{-27}= \\
16.7375 \times 10^{-27}= \\
1.67375 \times 10^{-26} \approx \\
1.67 \times 10^{-26}$

The answer is A. 1.67 × 10⁻²⁶ kg.

8 0

4 years ago