All categories

3 years ago

In order from LEAST density to most in order

Chemistry

1 answer:

baherus [9]3 years ago

4 0

Ping pong ball - 0.0840 g
Bottle cap - 0.92 g
Marble - 2.711 g

You might be interested in

n the 1990’s, the Boeing Co. developed a process for treating wastewater from electroplating and printed circuit board manufactu

scZoUnD [109]

Answer:

Final concentrations:

Cu²⁺ = 0

Al³⁺ = 3.13 mmol/L = 84.51 mg/L

Cu = 4.7 mmol/L = 300 mg/L

Al = 0.57 mmol/L = 15.49 mg/L

Explanation:

2Al (s) + 3Cu²⁺ (aq) → 2Al³⁺ (aq) + 3Cu (s)

Al: 27 g/mol ∴ 100 mg = 3.7 mmol

Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol

3 mol Cu²⁺ _______ 2 mol Al

4.7 mmol Cu²⁺ _____ x

x = 3.13 mmol Al

4.7 mmol of Cu²⁺ will be consumed.

3.13 mmol of Al will be consumed.

4.7 mmol of Cu will be produced.

3.13 mmol of Al³⁺ will be produced.

0.57 mmol of Al will remain.

6 0

2 years ago

which of the following cannot be considered a single phase? a) a pure solid b) a pure liquid c) a homogeneous mixture d) a heter

Trava [24]

The question is asking to choose among the following choices is cannot be considered as a single phase and base on my further research and understanding about the sad topic, I would say that the answer would be <span>d) a heterogeneous mixture. I hope you are satisfied with my answer </span>

8 0

3 years ago

What is the volume of an object with a density of 2.6 g/cm³ and a mass of 30.5g?

inessss [21]

Answer:

The answer is

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

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

From the question

mass of object = 30.5 g

Density = 2.6 g/cm³

The volume is

$volume = \frac{30.5}{2.6} \\ = 11.7307692...$

We have the final answer as

Hope this helps you

5 0

3 years ago

5. How much heat (in calories) is absorbed by a reaction when

Wewaii [24]

Answer:

1.2 × 10⁴ cal

Explanation:

Given data

Mass of water (m): 300 g

Initial temperature: 80 °C

Final temperature: 40°C

We can calculate the heat released by the water ( $Q_w$ ) when it cools using the following expression.

$Q_w = c \times m \times (T_f - T_i)$

where

c is the specific heat capacity of water (1 cal/g.°C)

$Q_w = \frac{1cal}{g.\°C} \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal$

According to the law of conservation of energy, the sum of the heat released by the water ( $Q_w$ ) and the heat absorbed by the reaction ( $Q_r$ ) is zero.