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Pachacha [2.7K]
3 years ago
6

A student measures the mass of an 8 cm block of brown sugar to be 12.9 g. What is the density of

Chemistry
1 answer:
horrorfan [7]3 years ago
8 0

Answer:

0.0025g/cm³

Explanation:

Given parameters:

Mass of the brown sugar = 12.9g

dimension of the sugar = 8cm

unknown:

density of the sugar = ?

Solution

Density is defined as the mass per unit volume of a substance. The expression is given below:

      Density = \frac{mass}{volume}

We know the mass of the substance but the volume is unknown:

Volume of the brown sugar = l x bx h = 8x8x8 = 512cm³

Density of the brown sugar = \frac{12.9}{512} = 0.0025g/cm³

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Box 1
Alex73 [517]

10g

Explanation:

Box 1, Mass of A = 10g

Box 2, Mass of B = 5g

Box 3, = 1A + 1B

Unknown:

Mass of B that would combine with mass of 20g of A

Solution:

   Mass ratio of A to B:

   \frac{mass of A}{mass of B} = mass ratio

           \frac{10}{5} = mass ratio

        The mass ratio of A to B = 2: 1

Now, number of B that will combine with 20g of A;

       

           \frac{mass of A}{mass of B} = mass ratio

               \frac{20}{mass of B} = \frac{2}{1}

                     Mass of B = 10g

10g of B would combine with 20g of A

learn more:

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7 0
3 years ago
Read 2 more answers
What is the percent composition of NaHCO3?
zhuklara [117]

Answer:

                Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

              a) Na  =  22.99 g/mol

             b) H  =  1.01 g/mol

              c) C  =  12.01 g/mol

              d) O₃  =  16.0 × 3 =  48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

              Na  =  22.99 g/mol

             H    =  1.01 g/mol

              C    =  12.01 g/mol

              O₃  =  48 g/mol

                       ----------------------------------  

Total                  84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

                 =  22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  27.36 %

For H:

                 =  1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  1.20 %

For C:

                 =  12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  14.29 % ≈ 14.30 %

For O:

                 =  48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  57.13 % ≈ 57.14 %

8 0
3 years ago
A first order reaction has a rate constant of 0. 543 at 25°C. Given that the activation energy is 75. 9 kj/mol. Calculate the ra
Rasek [7]

The rate constant of first order reaction at 32. 3 °C is 0.343 /s must be less the 0. 543 at 25°C.

First-order reactions are very commonplace. we have already encountered  examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to present t-butanol. every other reaction that famous obvious first-order kinetics is the hydrolysis of the anticancer drug cisplatin.

The value of ok suggests the equilibrium ratio of products to reactants. In an equilibrium combination both reactants and merchandise co-exist. big ok > 1 merchandise are k = 1 neither reactants nor products are desired.

Rate constant K₁  = 0. 543 /s

T₁  = 25°C

Activation energy Eₐ =  75. 9 k j/mol.

T₂ = 32. 3 °C.

K₂ =?

formula;

log K₂/K₁=  Eₐ /2.303 R [1/T₁ - 1/T₂]

putting the value in the equation  

K₂ = 0.343 /s

Hence, The rate constant of first order reaction at 32. 3 °C is 0.343 /s

The specific rate steady is the proportionality consistent touching on the fee of the reaction to the concentrations of reactants. The fee law and the specific charge consistent for any chemical reaction should be determined experimentally. The cost of the charge steady is temperature established.

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4 0
1 year ago
What are mountins ussaly made from
Katarina [22]

Answer:

rocks and earth

Explanation:

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5 0
3 years ago
Read 2 more answers
1) A 10g sample of H2(g) reacts with a 22g sample of O2(g) according to
solmaris [256]

Answer:

H₂ is excess reactant and O₂ the limiting reactant

Explanation:

Based on the chemical reaction:

2H₂(g) + O₂(g) → 2H₂O

<em>2 moles of H₂ react per mole of O₂</em>

<em />

To find limiting reactant we need to convert the mass of each reactant to moles:

<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>

10g H₂ * (1mol / 2.016g) = 4.96 moles

<em>Moles O₂ -Molar mass: 32g/mol-:</em>

22g O₂ * (1mol / 32g) = 0.69 moles

For a complete reaction of 0.69 moles of O₂ are needed:

0.69mol O₂ * (2mol H₂ / 1mol O₂) = 1.38 moles of H₂

As there are 4.96 moles,

<h3>H₂ is excess reactant and O₂ the limiting reactant</h3>
7 0
3 years ago
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