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miskamm [114]
3 years ago
13

5.1 ounces to grams.

Chemistry
2 answers:
Sergeeva-Olga [200]3 years ago
6 0

Answer: here’s the answer

Explanation:5.1 ounces =

144.583 grams

Formula:

for an approximate result, multiply the mass value by 28.35

Firdavs [7]3 years ago
5 0

Answer:

144.583 grams

Explanation:

thats what i found

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PLEASE HELP!! ASAP!! WILL MARK BRAINLIEST!!!
sdas [7]

Answer:

Alkali metals are soft and have low melting points.

7 0
3 years ago
Given the balanced equation:
Aliun [14]

Answer:

0.4

Explanation:

Given data:

Number of moles of SrCl₂ consumed = ?

Mass of ZnCl₂ produced = 54 g

Solution:

Chemical equation:

ZnSO₄ + SrCl₂    →    SrSO₄ +  ZnCl₂

Number of moles of ZnCl₂:

Number of moles = mass/ molar mass

Number of moles = 54 g/136.3 g/mol

Number of moles = 0.4 mol

Now we will compare the moles of  ZnCl₂ with SrCl₂  from balance chemical equation.

                          ZnCl₂              :             SrCl₂

                              1                  :                1

                           0.4                 :              0.4

Thus when 54 g of  ZnCl₂ produced 0.4 moles of SrCl₂ react.

6 0
3 years ago
Draw the major organic substitution product(s) for (2R,3S)-2-bromo-3-methylpentane reacting with the given nucleophile. Indicate
Andrew [12]

Answer:

(2R,3S)-2-ethoxy-3-methylpentane

and

(2S,3S)-2-ethoxy-3-methylpentane

Explanation:

For this case, we will have  CH_3CH_2O^- as nucleophile. Also, this compound is also in excess. So, we will have as solvent CH_3CH_2OH a protic solvent. Therefore the Sn1 reaction would be favored.

The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).

7 0
3 years ago
What is the pressure of 0.5 mol of nitrogen gas in a 5 L container at 203 K
vlada-n [284]

Answer:

=1.666 liters

Explanation:

1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.

0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1

=11.2 liters.

Standard pressure= 1 atmosphere (Atm)

Standard temperature = 273.15 Kelvin

According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂

Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.

Therefore P₂ =P₁V₁T₂/T₁V₂

Substituting for the values we get:

P₂= (1 atm× 11.2L ×203K)/ (273K×5L)

=1.666 atm

5 0
3 years ago
Nevermind, scratch the question
Lyrx [107]
What question there’s no question on here
5 0
2 years ago
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