Which of these choices best describes a pedigree?

Calculate the normality of a solution containing 147 g of h2s04 in 2L of solution

babunello [35]

I am not sure plz show me the question

5 0

3 years ago

A radioactive element reduces to 5.00% of its initial mass in

Stolb23 [73]

The half-life in months of a radioactive element that reduce to 5.00% of its initial mass in 500.0 years is approximately 1389 months

To solve this question, we'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Amount remaining (N) = 5%

Original amount (N₀) = 100%

<h3>Number of half-lives (n) =?</h3>

N₀ × 2ⁿ = N

5 × 2ⁿ = 100

2ⁿ = 100/5

2ⁿ = 20

Take the log of both side

Log 2ⁿ = log 20

nlog 2 = log 20

Divide both side by log 2

n = log 20 / log 2

Thus, 4.32 half-lives gas elapsed.

Finally, we shall determine the half-life of the element. This can be obtained as follow.

Number of half-lives (n) = 4.32

Time (t) = 500 years

t½ = t / n

t½ = 500 / 4.32

t½ = 115.74 years

Multiply by 12 to express in months

t½ = 115.74 × 12

<h3>t½ ≈ 1389 months </h3>

Therefore, the half-life of the radioactive element in months is approximately 1389 months

Learn more: brainly.com/question/24868345

8 0

3 years ago

If the density of a 45.0 cm3 block of wood is 0.65 g/ml what is the mass of the wood?

wlad13 [49]

Answer:

$\boxed {\tt mass=29.25 \ grams}$

Explanation:

The density formula is mass over volume.

$d=\frac{m}{v}$

Rearrange the formula for the mass, m. Multiply both sides of the formula by v.

$d*v=\frac{m}{v}*v$

$d*v=m$

Mass can be found by multiplying the density and volume. The density is 0.65 grams per milliliter and the volume is 45.0 cubic centimeters.

A cubic centimeter is equal to a milliliter.

Therefore, 45 cubic centimeters also equals 45 milliliters.

$d= 0.65 \ g/mL\\v= 45 \ mL$

Substitute the values into the formula.

$0.65 \ g/mL * 45 \ mL=m$

Multiply. Note the milliliters, or mL will cancel out.

$0.65 \ g * 45=m$

$29.25 \ g=m$

The mass of the wood is 29.25 grams.

5 0

3 years ago

Calculate the pH of 1.00 L of a buffer that contains 0.105 M HNO2 and 0.170 M NaNO2. What is the pH of the same buffer after the

kari74 [83]

Answer:

1.- pH =3.61

2.-pH =3.53

Explanation:

In the first part of this problem we can compute the pH of the buffer by making use of the Henderson-Hasselbach equation,

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the conjugate base anion concentration ( [NO₂⁻]), [HA] is the weak acid concentration,[HNO₂].

In the second part, our strategy has to take into account that some of the weak base NO₂⁻ will be consumed by reaction with the very strong acid HCl. Thus, first we will calculate the new concentrations, and then find the new pH similar to the first part.

First Part

pH = 3.40+ log {0.170 /0.105}

pH = 3.61

Second Part

# mol HCl = ( 0.001 L ) x 12.0 mol / L = 0.012

# mol NaNO₂ reacted = 0.012 mol ( 1: 1 reaction)

# mol NaNO₂ initial = 0.170 mol/L x 1 L = 0.170 mol

# mol NaNO₂ remaining = (0.170 - 0.012) mol = 0.158

# mol HNO₂ produced = 0.012 mol

# mol HNO₂ initial = 0.105

# new mol HNO₂ = (0.105 + 0.012) mol = 0.117 mol

Now we are ready to use the Henderson-Hasselbach with the new ration. Notice that we dont have to calculate the concentration (M) since we are using a ratio.

pH = 3.40 + log {0.158/.0117}

pH = 3.53

Notice there is little variation in the pH of the buffer. That is the usefulness of buffers.

4 0

3 years ago

Read 2 more answers

Name a possible product of this reaction in the presence of ether and AlCl3: methylbenzene + 1-chlorodecane.a. 1-methyl-2-decylb

Vikki [24]

Answer:

None of these

Explanation:

Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.

Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :

Step -1 : Generation of stable carbocation.

Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.

Step-2: Attack of the ring to the carbocation

The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.

The product formed is shown in mechanism does not mention in any of the options.

So, None of these is the answer

8 0

4 years ago

Read 2 more answers