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3 years ago

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Chemistry

1 answer:

Igoryamba3 years ago

8 0

Answer:

The arm that was not sprayed with anything

Explanation:

The control group would be the arm that was not sprayed with anything.

The control group during an experiment is a group that forms the baseline for comparison in other to determine the effects of a treatment. The control group does not include the variable that is being tested and as such, it provides the benchmark to measure the effects of the tested variable on the other group - the experimental group. In this case, the experimental group would be the arm that was sprayed with the repellent.

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A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na

s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f: after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

$Normality=molarity \times n-factor$

but in case of NaOH and HCl n-factor is 1 for each.

hence

normality=molarity;

At equivalence point: $N_1V_1=N_2V_2$

Before equivalence point : $N_1V_1>N_2V_2$

After the equivalence point: $N_1V_1$

$N_1V_1=100\times1=100$

case a: 5.00 mL of 1.00 M NaOH

$N_2V_2=5\times1=5$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

$N_2V_2=100\times1=100$

$N_1V_1=N_2V_2$ hence it is equivalence point

case c: 10.0 mL of 1.00 M NaOH

$N_2V_2=10\times1=10$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

$N_2V_2=150\times1=150$

$N_1V_1$ hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

$N_2V_2=50\times1=50$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

$N_2V_2=200\times1=200$

$N_1V_1$ hence it is after the eqivalence point

7 0

3 years ago

Calculate the average weight of the dogs listed in the chart below.

pishuonlain [190]

Answer:

Explanation:

Once you add all the weights together, you end up with 114 then you divide it by how many dogs there are, 6. 114/6 = 19. The average is 19 lbs.

3 0

3 years ago

Converting 53.3g of oxygen to moles will give you???

Marat540 [252]

Answer:

3.33 mol

Explanation:

1 g of oxygen is 0.062502343837894 mol

53.3 * 0.062502343837894

3.3313749265597505 mol

8 0

3 years ago

Deprenyl is an enzyme inhibitor that helps prevent the metabolism of dopamine in the brain. The chemical formula of Deprenyl is

mel-nik [20]

Answer:

0.007 g of deprenyl dose is required fro the patient with body mass of 70 kilograms.

Explanation:

The dose for treating Parkinson’s disease = 100 μg/kg body weight

Mass of patient's body = 70 kg

Amount of dose of deprenyl required = 100 μg/kg × 70 kg = 7,000 μg

1 μg = 0.00001 g

7,000 μg = 7,000 × 0.000001 g = 0.007 g

0.007 g of deprenyl dose is required fro the patient with body mass of 70 kilograms.

6 0

3 years ago

After 11.5 days, 12.5% of a sample of radon-222 that originally weighed 42g remains. What is the half-life of this isotope?

Bond [772]

Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value. The equation to describe the decay is
Nt=N0(1/2) $^{t/t(1/2)}$ where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time. So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days

5 0

3 years ago