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allochka39001 [22]

4 years ago

14

A botanist is measuring plants to determine amount of growth after fertilizer was used. A ruler was used for the measurements. W

hich type of measurement is being collected?

1 answer:

luda_lava [24]4 years ago

3 0

Quantitative measurement

Explanation:

Using the ruler to measure the height of the plant is a quantitative measurement. Quantitative measurement is a from of measurement that can be measured and numbers affixed.

The plant growth is the dependent variable in this experimental setup since it is the effect produced by applying fertilizers to the plant.
Using the ruler the botanist can tell about the growth of the plant in numbers and not by mere qualitative terms.
Other examples of quantitative measurements are: a mass of broth, height of an animal e.t.c.

Learn more:

numbers brainly.com/question/154071

#learnwithBrainly

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Iodine and Calcium is the correct answer

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Given that this reaction is exothermic, what direction will the equilibrium shift when the temperature of the reaction is decrea

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Equilibrium will shift towards the products when temperature is decreased in an exothermic reaction of the formation of ammonia.

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An exothermic reaction is a reaction in which heat content of the reactants is greater than the heat content of product.

In an exothermic reaction, heat is given off.

For an exothermic reaction in equilibrium, increasing temperature shifts equilibrium to the towards the left, towards the reactants.

On the other, equilibrium will shift towards the products when temperature is decreased.

Therefore, equilibrium will shift towards the products when temperature is decreased in the reaction of the formation of ammonia.

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2 years ago

What is the general trend for electron affinity values going across a period? The

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3 years ago

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44.8% of a 250. mL acid solution is used in an experiment, what volume of acid (in mL) was used?

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Answer:

112 mL

Explanation:

The formula for percent by volume is

$\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%$

If you have 250 mL of a solution that is 44.8 % v/v,

$\begin{array}{rcl}44.8\, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\44.8 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{44.8 \times 250\text{ mL}}{100}\\\\ & = & \textbf{112 mL}\\\end{array}$

7 0

3 years ago

Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be

AlexFokin [52]

Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

$\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles$

$\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles$

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

$\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles$

The concentration of benzoic acid is:

$C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M$

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺

0.14 - x x x

$Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}$

$Ka = \frac{x*x}{0.14 - x}$

$6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0$ (2)

By solving equation (2) for x we have:

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

$pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52$

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻ (3)

The number of moles of C₆H₅CO₂⁻ is:

$\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles$

The volume of NaOH added is:

$V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L$

The concentration of C₆H₅CO₂⁻ is:

$C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M$

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻

0.14 - x x x

$Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}$

$(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0$

$(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0$

By solving the equation above for x, we have:

x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

The pH is:

$pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33$

$pH = 14 - pOH = 14 - 5.33 = 8.67$

c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:

$\eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles$

From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:

$\eta_{NaOH} = \eta_{NaOH}i - \eta_{C_{6}H_{5}CO_{2}H} = 0.025 moles - 0.015 moles = 0.010 moles$

The concentration of NaOH is:

$C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M$

Therefore, the pH is given by this excess of NaOH:

$pOH = -log([OH^{-}]) = -log(0.067) = 1.17$

$pH = 14 - pOH = 12.83$

I hope it helps you!

4 0

3 years ago