All categories

3 years ago

Stronger acids have more _______

1 answer:

7 0

Examples of strong acids are hydrochloric acid (HCl), perchloric acid (HClO4), nitric acid (HNO3) and sulfuric acid (H2SO4). ... For example, hydrogen chloride is a strong acid in aqueous solution, but is a weak acid when dissolved in glacial acetic acid.

You might be interested in

The relatively small, rocky bodies generally found orbiting between mars and jupiter are known as _____. meteoroids satellites c

adelina 88 [10]

The answer that is being described above is the ASTEROIDS. The one that we see floating between Mars and Jupiter is what we call the Asteroid Belt. The asteroid belt comprises of different rocky bodies and they also orbit within the solar system. Hope this helps.

6 0

2 years ago

Read 2 more answers

Reducing, reusing, and recycling in your office is likely to _______.

iris [78.8K]

Conserve natural resources, energy and landfill space.

4 0

3 years ago

A tank with a volume of 0.150 m3 contains 27.0oC helium gas at a pressure of 100 atm. How many balloons can be blown up if each

jekas [21]

Answer:

884 balloons

Explanation:

Assume ideal gas, since temperature is constant, then the product of pressure and volume is constant.

So if pressures reduces from 100 to 1.2, the new volume would be

$V_2 = \frac{P_1V_1}{P_2} = \frac{100*0.15}{1.2} = 12.5 m^2$

The spherical volume of each of the balloon of 30cm diameter (15 cm or 0.15 m in radius) is

$V_b = \frac{4}{3}\pir^3 = \frac{4}{3}\pi 0.15^3 = 0.014 m^3$

The number of balloons that 12.5 m3 can fill in is

$V_2/V_b = 12.5 / 0.014 = 884$

8 0

2 years ago

If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

4 0

3 years ago

A pulled tablecloth exerts a frictional force of 0.6 n on a plate with a mass of 0.4 kg. what is the acceleration of the plate?

Ludmilka [50]

The formula for acceleration is a = F/m; Where: F = force; m = mass
Given: F = .6n; m = .4kg; a = ?
a = F/m
= .6/.4
= 1.5

Therefore, the acceleration of the plate is 1.5 m/s^2

6 0

3 years ago