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3 years ago

A 133.8-gram sample of bronze was 10.3% tin by mass. Determine the total mass of tin in the sample.

1 answer:

6 0

You just need to multiply the total mass by the decimal value of the part that is tin. 133.8*0.103=13.8g (following the rules of significant figures).

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Find concentration of solution of 45g of glucose is dissolved in water to prepare 500g solution?

ivolga24 [154]

Answer:

9% solution by mass

Explanation:

If there is 500 gm of solution and 45 g of it is glucose then:

45/500 * 100% = 9 % solution

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2 years ago

URGENTTTTT What is the correct formula when Al+3 combines with CrO4-2

PIT_PIT [208]

A and C are incorrect because they are not complete transfer of valence electrons. Ionic bonds best to form a neutral molecule

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2 years ago

An astronaut is free-floating in space near a space station and it is not tethered to anything. He has some tools and a fire ext

worty [1.4K]

Hey there

the answers are

and

D. He could spray the fire extinguisher in the opposite direction of the space station

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3 years ago

Read 2 more answers

What determines the traits of offspring?A.Food sources that have been genetically engineered B.Literary metaphors and exciting c

iren2701 [21]

Answer:

D. Genes received from the offsprings parents.

Explanation:

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2 years ago

The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make

natta225 [31]

Answer:

the energy of the third excited rotational state $\mathbf{E_3 = 16.041 \ meV}$

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = $\dfrac{m_1 \times m_2}{m_1 + m_2}$

μ = $\dfrac{1 \times 35}{1 + 35}$

μ = $\dfrac{35}{36}$

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ = $\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg$

μ = 1.6139 × 10⁻²⁷ kg

$r_o = 127 \ pm = 127*10^{-12} \ m$

The rotational level Energy can be expressed by the equation:

$E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)$

where ;

J = 3 ( i.e third excited state) &

$I = \mu r^2_o$

$E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)$

$E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)$

$E_3= 2.5665 \times 10^{-21} \ J$

We know that :

1 J = $\dfrac{1}{1.6 \times 10^{-19}}eV$

$E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV$

$E_3 = 16.041 \times 10 ^{-3} \ eV$

$\mathbf{E_3 = 16.041 \ meV}$

8 0

2 years ago