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3 years ago

BRAINLIESTTT ASAP!!! PLEASE HELP ME :))

Chemistry

1 answer:

garri49 [273]3 years ago

6 0

Answer:

Multiply 1.25 by 0.04 and divide the result obtained by 1,000

Explanation:

Given: [1 gram = 0.04 ounce, 1 liter = 1,000 milliliter]

1.25 x 0.04 = 0.05 oz

Therefore, 0.05 per 1,000 milliliter

0.05 ÷ 1,000 = 0.00005 oz

Therefore, the density of the gas is 0.00005 oz/mL

Hope this helps! :)

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The initial concentration of reactant in a first-order reaction is 0.27 m. the rate constant for the reaction is 0.75 s-1. what

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The concentration in mol/l of reactant after 1.5 s is calculated as follows

from first order integrated equation

In (A)t = - Kt + In (A)o where
At = final concentration =?
A)o =initial concentration =0.27 M
K = constant=0.75
T=temperature = 1.5 s

in( A)t = -0.75(1.5) + in(0.27)

In (A)t =-1.125 +-1.31 =-2.435
In(A)t =-2.435

In (A)t = e
find the e value

( A)t is therefore =0.0876 mol/l

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4 years ago

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Which word means the opposite of boiling melting

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Freezing could be a good answer

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3 years ago

Considering that both sodium (Na) and cesium (Cs) are both alkali metals, which of the following is a correct description of the

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Answer:- C) sodium has a higher ionization energy because it is smaller.

Explanations:- Ionization energy is the energy required to remove electron from valence shell of an atom in it's gaseous form.

More energy is required to remove the electron if the size of the atom is smaller as it's electrons are more strongly attracted by it's nucleus.

In periodic table, atomic size decreases on moving left to right in a period and increases on going top to bottom in a group.

Na and Cs are the elements of same group(first group) and also Cs is below Na. So, as Na is placed above Cs, the size of Na is smaller than Cs and so more energy is required to remove the electron from this smaller atom.

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4 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),

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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

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

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