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3 years ago

Jermey has two plants. he waters both equally. he puts one plant outside. he puts another plant in a closet

Chemistry

2 answers:

storchak [24]3 years ago

5 0

Answer:

A. The plant in the closet will die

- A plant needs water, soil and sunlight to grow.

Ivanshal [37]3 years ago

3 0

The plant in the closet will die without the proper nutrients granted through sunlight

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How many moles of nitrous oxide gas (N2O) is contained in 20.0 L balloon at a pressure of 1.09 atm and a temperature of 298K? Ro

prohojiy [21]

Answer:

There are 0,89 moles of nitrous oxide gas in the balloon.

Explanation:

We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol:

PV= nRT ---> n= PV/RT

n= 1,09 atm x 20,0 L /0.082 l atm / K mol x 298 K

n= 0,89212637 mol

7 0

3 years ago

If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),

garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] (1)

Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻

pH = 7.40

pKa = 7.20

[A-] + [HA] = 0.10moles (2)

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

4 0

3 years ago

In the reaction Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq), how many moles of hydrogen gas will be produced from 250.0 milliliters of

satela [25.4K]

Mg + 2HCl = H₂ + MgCl₂

n(HCl)=c(HCl)v(HCl)

n(H₂)=n(HCl)/2=c(HCl)v(HCl)/2

n(H₂)=3.0mol/L*0.2500L/2=0.375 mol≈0.38 mol

5 0

3 years ago

Read 2 more answers

How do electrons populate atoms?

lutik1710 [3]

Electrons fill the electron orbitals (s, p, d, or f) starting from the lowest energy level going to the highest energy level. I hope this helps. let me know if you need more information such as what the

7 0

3 years ago

8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a

Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

$C_{1} V_{1} =C_{2} V_{2}$

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, $(5.45*V_{1} ) = (820.7*1.41)$

$V_{1}=\frac{820.7*1.41}{5.45}= 212.33 ml$

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

3 0

3 years ago