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3 years ago

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Becky wanted to figure out what type of work would work the best for growing beans see you watered one with Coca-Cola one with l

emonade and another one just with water after one week she measure how high they had a growth (IV) (DV) (CONTROL)

1 answer:

Lubov Fominskaja [6]3 years ago

8 0

IV: type of liquid used to water the plant (coca-cola, lemonade, water)
DV: height of growth
Control: time grown, same temperature and location

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What is the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm?

Volgvan

Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

Explanation:

Given: $V_{1}$ = 61 L, $T_{1}$ = 183 K, $P_{1}$ = 0.60 atm

At STP, the value of pressure is 1 atm and temperature is 273.15 K.

Now, formula used to calculate the new volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{0.60 atm \times 61 L}{183 K} = \frac{1 atm \times V_{2}}{273.15 K}\\V_{2} = 54.63 L$

Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

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2 years ago

Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is

Citrus2011 [14]

Depression in freezing point (Δ $T_{f}$ ) = $K_{f}$ ×m×i,
where, $K_{f}$ = cryoscopic constant = $1.86^{0} C/m$ ,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For $NaNO_{3}$ )

Thus, (Δ $T_{f}$ ) = 1.86 X 0.0085 X 2 = $0.03162^{0}C$

Now, (Δ $T_{f}$ ) = $T^{0}$ - T
Here, T = freezing point of solution
$T^{0}$ = freezing point of solvent = $0^{0}C$
Thus, T = $T^{0}$ - (Δ $T_{f}$ ) = - $0.03162^{0}C$

8 0

3 years ago

Read 2 more answers

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,

<u>pH = 14 - 4.64 = 9.36</u>

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3 years ago

Zinc, sulphur, oxygen<br> What would this compound be called?

Nataliya [291]

Zinc sulphate

Zinc sulphateZnSO₄

This is the answer.

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2 years ago

How many electrons occupy a filled 7s sublevel

Natali5045456 [20]

Answer:f has 14 electrons in 7 sublevel orbitals,d has 10 electrons in 5 sublevel orbitals,p has 6 electrons in 3 sublevel orbitals,s has 2 electrons in 1 sublevel orbital.

Explanation:

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3 years ago

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