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Aloiza [94]
2 years ago
11

Pls help and match them

Chemistry
1 answer:
slega [8]2 years ago
3 0

Answer:

Simile: C

Personification: D

Symbol: A

Hyperbole: B

Metaphor: E

Explanation:

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A 1.20-L container contains 1.10 g of an unknown gas at STP. What is the molecular weight of the unknown gas?
SOVA2 [1]

Answer:

M = 20.5 g/mol

Explanation:

Given data:

Volume of gas = 1.20 L

Mass of gas = 1.10 g

Temperature and pressure = standard

Solution:

First of all we will calculate the density.

Formula:

d = mass/ volume

d = 1.10 g/ 1.20 L

d = 0.92 g/L

Now we will calculate the molar  mass.

d = PM/RT

0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K

M =  0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K /  1 atm

M = 20.5 g/mol

8 0
2 years ago
3. What methods are you using to test this hypothesis? Outline the steps of the procedure in full sentences.
Ulleksa [173]

Answer:

AHYPOTHISES IS A GUESS

Explanation:

4 0
2 years ago
In 3–5 sentences, explain why electrical wires are often constructed from metal coated with rubber, and how that is similar to m
stich3 [128]

Answer:

Covering electrical wires in plastic guarantees that the electrons flowing through the wires will not flow through your body when the wire is touched.

Explanation:

"The purpose of insulation covering the metal part of an electrical wire is to prevent accidental contact with other conductors of electricity, which might result in an unintentional electric current through those other conductors."

https://www.allaboutcircuits.com/worksheets/wire-types-and-sizes/#:~:text=Most%20electrical%20wire%20is%20covered,or%20plastic%20coating%20called%20insulation.&text=The%20purpose%20of%20insulation%20covering,current%20through%20those%20other%20conductors.

https://www.hunker.com/13414152/why-are-electrical-wires-covered-in-plastic

7 0
2 years ago
A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A
Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

4 0
3 years ago
All of the following are forms of water in the atmosphere EXCEPT _________.
solong [7]

Answer:

OZONE

Explanation:

HOPE IT HELP

MARK ME AS BRAINLIEST

3 0
2 years ago
Read 2 more answers
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