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wel
1 year ago
8

How many quarters fit in a 1.75 liter bottle?

Chemistry
2 answers:
WARRIOR [948]1 year ago
8 0
1.85 quarts can fit into a 1.75 liter bottle
Volgvan1 year ago
3 0

Answer:

In a 1.75 liter bottle fit 1.8492 quarters.

Explanation:

The quarter is a unit of volume. It is equivalent to \frac{1}{4} of a gallon :

1gallon=3.78541L

Where L is liter

⇒

\frac{1}{4}gallon=1quarter=\frac{3.78541L}{4}

To find how many quarters fit in a 1.75 L bottle we can write :

\frac{\frac{3.78541L}{4}}{1quarter}=\frac{1.75L}{x}

Solving for x :

x=\frac{(1.75L).(1quarter)}{\frac{3.78541L}{4}}

x=1.8492quarters

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The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
The chemical equation, Cr + Fe(NO3)2 → Fe + Cr(NO3)3, is an example of which type of reaction?
Angelina_Jolie [31]

Answer:

Redox type

Explanation:

The reaction is:

2Cr +  3Fe(NO₃)₂ → 2Fe + 2Cr(NO₃)₃

2 moles of chromium can react to 3 moles of iron (II) nitrate in order to produce 2 moles of iron and 2 moles of chromium nitrate.

If we see oxidation state, we see that chromium changes from 0 to +3

Iron changed the oxidation state from +2 to 0

Remember that elements at ground state has 0, as oxidation state.

Iron is being reduced while chromium is oxidized. Then, the half reactions are:

Fe²⁺  +  2e⁻ ⇄  Fe    (Reduction)

Cr ⇄ Cr³⁺  +  3e⁻    (Oxidation)

When an element is being  reduced, while another is being oxidized, we are in prescence of a redox reaction.

8 0
2 years ago
How many Ca2+ ions are in a sample of CaSO4 having a total mass of 68.07 g?
harkovskaia [24]

Answer:

There are  

17.01

Explanation:

The chemical formula for calcium phosphate is  

Ca

3

(PO

4

)

2

. This means that in one mole of calcium phosphate, there are three calcium ions and two phosphate ions.

6 0
1 year ago
Write an equation that represents the action in water of rubidium hydroxide as an Arrhenius base.
Anika [276]

Answer:

RbOH  → Rb⁺ +  OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

Explanation:

Arrhenius theory states that a compound is considered a base, if the compound can generate OH⁻ ions in aqueous solution.

Our compound is the RbOH.

When it is put in water, i can dissociate like this:

RbOH  → Rb⁺ +  OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

3 0
2 years ago
Enter your answer in the provided box. An industrial chemist introduces 2.0 atm of H2 and 2.0 atm of CO2 into a 1.00−L container
PSYCHO15rus [73]

Answer: 0.0944 gram of H2

Explanation:

Raising the T from 25 C (298 K) to 700 C (973 K) increases the pressure of each gas by:

2.0 atm x (973 K / 298 K) = 6.53 atm

Where

Kc = Kp because the moles of product equals the moles of reactants.

At equilibriuim, the amounts are

P(H2) = 6.53 - x

P(CO2) = 6.53 - x

P(H2O) = x

P(CO) = x

Kc = Kp = .534 = (x)(x) / [(6.53 - x)(6.53 - x)]

Take the square root of each side

(.534)^0.5 = x / (6.53 - x)

x = 0.731 (6.53 - x)

x = 4.77 - 0.731x

1.731x = 4.77

x = 4.77 / 1.731 = 2.76 atm

P(H2) at equilibriuim = 6.53 - 2.76 = 3.77 atm

P(CO2) at equilibrium = 6.53 - 2.76 = 3.77 atm

PV = nRT

n = PV/RT = [(3.77 atm)(1.00 L)] / [(0.08206 L atm/K mol)(973 K)] = 0.0472 mol H2

0.0472 mol H2 x (2.00 g / 1.00 mol) = 0.0944 g

8 0
2 years ago
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