All categories

2 years ago

Calculate the density of an aluminum block with a volume of 100 cm3 and a mass of 270g.

Chemistry

1 answer:

m_a_m_a [10]2 years ago

6 0

Answer:

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{270}{100} = \frac{27}{10} \\$

We have the final answer as

Hope this helps you

You might be interested in

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin

erastovalidia [21]

Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

8 0

2 years ago

I need help with this, please :00000

mojhsa [17]

theoretical yield of the reaction is 121.38 g of NH₃ (ammonia)

limiting reactant is N₂ (nitrogen)

excess reactant is H₂ (hydrogen)

Explanation:

We have the following chemical reaction:

N₂ + 3 H₂ → 2 NH₃

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

From the chemical reaction we see that 3 moles of H₂ are reacting with 1 moles of N₂, so 50 moles of H₂ are reacting with 16.66 moles of N₂ but we only have 3.57 moles of N₂ available, so the limiting reactant will be N₂ and the excess reactant will be H₂.

Knowing the chemical reaction and the limiting reactant we devise the following reasoning:

if 1 mole of N₂ produce 2 moles of NH₃

then 3.57 moles of N₂ produce X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g

theoretical yield of the reaction is 121.38 g of NH₃

Learn more about:

limiting reactant

brainly.com/question/13979150

#learnwithBrainly

4 0

3 years ago

1. power utilizes moving water to generate electricity.

Marianna [84]

Hydroelectric energy

8 0

3 years ago

*URGENT*

DaniilM [7]

Answer:

d hope this helps u hhwjs GD jehehj

4 0

3 years ago

Read 2 more answers

snow_lady [41]

The suggestion is to prevent a puddle of the liquid present in the sample from forming or from it leaking on to the surface on which it is placed. For example, if precipitates of a solid are removed from water and then placed on filter paper to dry, the water will soak into the filter paper and then leak on to the counter on which it is placed. If this precipitate were placed in a watch glass or weighing paper, the water would only evaporate and would not contaminate the sample.

5 0

3 years ago