All categories

3 years ago

A block weighing 13.15 grams has sides of 11.5 cm, 5.75 cm, and 8.44 cm. What is the density of the block in g/cm?

Chemistry

1 answer:

vitfil [10]3 years ago

7 0

Answer:

B. or .0236

Explanation:

Multiply the 3 sides to get 558.095, then divide the mass and volume to get .0236

You might be interested in

Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)

Sedbober [7]

Answer:

The yield of the product in gram is $\mathsf{{w_P}=0.26 \ gram}$

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = $2.7 \times 10^{-3} \ moles$

We know that:

Number of moles = mass/molar mass

Then:

$2.7 \times 10^{-3} = \dfrac{ mass}{257.997}$

$mass = 2.7 \times 10^{-3} \times 257.997$

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = $\dfrac{n_P}{n_R}\times 100\%$

where;

$n_P = \dfrac{w_P}{m_P}$ and $n_R = \dfrac{w_R}{m_R}$

Theoretical yield = $\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

Given that the theoretical yield = 100%

Then:

$100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

$\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}$

${w_P}=\dfrac{w_R \times m_P}{m_R}$

where,

$w_P$ = derived weight of the product

$m_P =$ the molecular mass of the derived product

$m_R =$ the molecular mass of the reagent

$w_R$ = weight in a gram of the reagent

${w_P}=\dfrac{w_R \times m_P}{m_R}$

${w_P}=\dfrac{0.697 \times 96.2}{257.997}$

$\mathsf{{w_P}=0.26 \ gram}$

8 0

3 years ago

PLEASE HELP HELP I NEED THIS ASAP

Sergeu [11.5K]

Wait SORRY HELP WITH WHAT?!!

5 0

3 years ago

Read 2 more answers

For 520.0 ml of a buffer solution that is 0.175 m in hc2h3o2 and 0.165 m in nac2h3o2, calculate the initial ph and the final ph

alisha [4.7K]

KWIJRJHO0907 X 023KO

3 0

3 years ago

Which factor is generally responsible for high melting points?

kherson [118]

The factor that is generally responsible for higher melting point is intermolecular forces. The compounds that are covalent in nature are made of molecules rather than ions. It has been seen that some of the covalent compounds have polar molecules at one end, due to which the one end has more electronegative force than the other. The electrostatic force that is bounding the compound is the main cause of higher melting point of this compound. So it is true that with the increase of polarity of a compound creates higher melting point. .. hope I helped

3 0

3 years ago

Which of the following represent a mixture?

7nadin3 [17]

The answer is B NaCI solid

6 0

3 years ago