Which surface is most likely to create an echo?

Chemistry

2 answers:

borishaifa [10]3 years ago

6 0

Answer:

a room with hard, bumpy walls

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velikii [3]3 years ago

6 0

The answer is A. if a sound meets a soft surface like C or D they will be absorbed and wouldn’t bounce back. the answer would have to be either A or B. But the answer is A because the wave remains more intact when it hits a smooth surface.

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Element Y has two natural isotopes Y-63 (62.940 amu) and Y-65 (64.928 amu). Calculate the atomic mass of element Y, given the ab

djverab [1.8K]

Answer:

Average atomic mass = 63.553 amu.

Explanation:

Given data:

Abundance of Y-63 = 69.17%

Abundance of Y-65 = 100 - 69.17 = 30.83%

Atomic mass of Y-63 = 62.940 amu

Atomic mass of Y-65 = 64.928 amu

Atomic mass of Y = ?

Solution:

Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass= (62.940×69.17)+(64.928×30.83) /100

Average atomic mass = 4353.560 + 2001.730 / 100

Average atomic mass = 6355.29 / 100

Average atomic mass = 63.553 amu.

5 0

4 years ago

The table compares the frequency of some unknown electromagnetic waves,

Sedaia [141]

Answer:

Wave X

Explanation:

because the lower the frequency, the longer the wave length

5 0

2 years ago

Magnesium acetate can be prepared by a reaction involving 15.0 grams of iron(III) acetate with either 10.0 grams of Magnesium Ch

noname [10]

1) Chemical reaction:
2(CH₃COO)₃Fe + 3MgCrO₄ → Fe₂(CrO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgCrO₄) = 10,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol.
n(MgCrO₄) = m(MgCrO₄) ÷ M(MgCrO₄).
n(MgCrO₄) = 10 g ÷ 140,3 g/mol.
n(MgCrO₄) = 0,071 mol; limiting reagens.
From chemical reaction: n(MgCrO₄) : n((CH₃COO)₂Mg) = 3 : 3.
n((CH₃COO)₂Mg) = 0,071 mol.
m((CH₃COO)₂Mg) = 0,071 mol · 142,4 g/mol.
n((CH₃COO)₂Mg) = 10,11 g.

2) Chemical reaction:
2(CH₃COO)₃Fe + 3MgSO₄ → Fe₂(SO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgSO₄) = 15,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol; limiting ragens.
n(MgSO₄) = m(MgSO₄) ÷ M(MgSO₄).
n(MgSO₄) = 15 g ÷ 120,36 g/mol.
n(MgSO₄) = 0,125 mol; limiting reagens.
From chemical reaction: n(CH₃COO)₃Fe) : n((CH₃COO)₂Mg) = 2 : 3.
n((CH₃COO)₂Mg) = 0,064 mol · 3/2.
n((CH₃COO)₂Mg) = 0,096 mol.
m((CH₃COO)₂Mg) = 0,096 mol · 142,4 g/mol.
m((CH₃COO)₂Mg) = 13,7 g.

7 0

4 years ago

A mixture of three noble gases has a total pressure of 1.25 atm. The individual pressures exerted by neon and argon are 0.68

FinnZ [79.3K]

First of all, as you seen the gases are noble which means that will not react with each other and in this case each gas create individual pressure.

P $_{T}$ = total pressure