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3 years ago

An object on top of a building has a GPE of 23,048j and a mass of 39kg, What is the height of the object

Chemistry

1 answer:

kirill115 [55]3 years ago

3 0

Answer:

Explanation:

The height of the object can be found by using the formula

$h = \frac{p}{mg } \\$

where

p is the potential energy

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

$h = \frac{23048}{39 \times 10} = \frac{23048}{390} \\ = 59.0974...$

We have the final answer as

Hope this helps you

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A chemist must prepare of sodium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask ab

77julia77 [94]

Answer:

0.0400 g for the example given below.

Explanation:

pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.

By definition, $pH = -log[H_3O^+]$ .
NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation: $NaOH (aq)\rightarrow Na^+ (aq) + OH^- (aq)$ .
From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH: $[NaOH] = [OH^-]$ .
Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain: $[OH^-] = [NaOH] = \frac{n_{NaOH}}{V} = \frac{m_{NaOH}}{M_{NaOH}V}$ .
We also know that $pOH = 14.00 - pH = -log[NaOH]$ . Take the antilog of both sides: $10^{-pOH} = 10^{pH - 14.00} = [NaOH] = \frac{m_{NaOH}}{M_{NaOH}V}$ .
Solve for the mass of NaOH: $m_{NaOH} = 10^{pH - 14.00}\cdot M_{NaOH}\cdot V$ .

Now, let's say that pH is given as 12.00 and we use a 100-ml volumetric flask. Then we would obtain:

$m_{NaOH} = 10^{12.00 - 14.00}\cdot 39.997 g/mol\cdot 0.100 L = 0.0400 g$

7 0

3 years ago

Convert 150 g/L to the unit g/mL. 15,000 g/mL 15 g/mL 0.15 g/mL 0.0015 g/mL

fredd [130]

1 g/L ------- 0.001 g/mL
150 g/L ----- ?

150 x 0.001 / 1

= 0.15 g/mL

Answer C

6 0

3 years ago

Read 2 more answers

Hcl and nh3 react to form a white solid, nh4cl. if cotton plugs saturated with aqueous solutions of each are placed at the ends

IgorLugansk [536]

24.4 cm.

<h3>Explanation</h3>

HCl and NH₃ reacts to form NH₄Cl immediately after coming into contact. Where NH₄Cl is found is the place the two gases ran into each other. To figure out where the two gases came into contact, you'll need to know how fast they move relative to each other.

The speed of a HCl or NH₃ molecule depends on its kinetic energy.

$E_\text{k} = 1/2 \; m \cdot v^{2}$

Where

$E_\text{k}$ is the kinetic energy of the molecule,
$m$ its mass, and
$v^{2}$ the square of its speed.

Besides, the kinetic theory of gases suggests that for an ideal gas,

$E_\text{k} \propto T$

where $\text{T}$ its temperature in degrees kelvins. The two quantities are directly proportional to each other. In other words, the average kinetic energy of molecules shall be the same for any ideal gas at the same temperature. So is the case for HCl and NH₃

$E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3)$

$m(\text{HCl}) \cdot v^{2}(\text{HCl}) = E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)$

Where

$m(\text{HCl})$ , $v(\text{HCl})$ , and $E_\text{k}(\text{NH_3})$ the mass, speed, and kinetic energy of an HCl molecule;
$m(\text{NH}_3)$ , $v(\text{NH}_3)$ , and $E_\text{k}(\text{NH}_3)$ the mass, speed, and kinetic energy of a NH₃ molecule.

The ratio between the mass of an HCl molecule and a NH₃ molecule equals to the ratio between their molar mass. HCl has a molar mass of 35.45; NH₃ has a molar mass of 17.03. As a result, $m(\text{HCl}) = 36.45 / 17.03 \; m(\text{NH}_3)$ . Therefore:

$36.45 /17.03\; m(\text{NH}_3) \cdot v^{2}(\text{HCl}) = m(\text{HCl}) \cdot v^{2}(\text{HCl}) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)$

$36.45 /17.03\; v^{2}(\text{HCl}) = v^{2}(\text{NH}_3)$

$\sqrt{36.45 /17.03}\; v(\text{HCl}) = v(\text{NH}_3)$

The average speed NH₃ molecules would be $\sqrt{36.45/17.03} \approx 1.463$ if the average speed of HCl molecules $v(\text{HCl})$ is 1.

$\text{Time before the two gases meet} = \frac{\text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}$

$\text{Distance from the HCl end} = v(\text{HCl}) \times \text{Time before the two gases meet}\\\phantom{\text{Distance from the HCl end}} = v(\text{HCl}) \times \frac{ \text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}\\\phantom{\text{Distance from the HCl end}} = \frac{v(\text{HCl})}{v(\text{HCl}) + v(\text{NH}_3)} \times \text{Length of the Tube}\\\phantom{\text{Distance from the HCl end}} = \frac{1}{1 + 1.463} \times 60.0\; \text{cm} \\\phantom{\text{Distance from the HCl end}} = 24.4 \; \text{cm}$

8 0

3 years ago

Can someone help?

victus00 [196]

1. The answer should be combustion

2. Water is known to successfully dissolve both acids and bases, and water is considered a universal solvent.

3. The answer should be balanced.

Hope this helps :)

7 0

3 years ago

Read 2 more answers

What is the molality of a solution made by combining 5.0 g of CaCl2 with 500.0 g of water?

ankoles [38]

Molality is defined as the number of moles of solute dissolved in 1 kg of solvent.
To calculate molality, we need to calculate the number of moles of CaCl₂.
Mass of CaCl₂ - 5.0 g
Molar mass of CaCl₂ - 111 g/mol
The number of moles of CaCl₂ - 5.0 g / 111 g/mol = 0.045 mol
we need to then calculate the number of moles in 1 kg solvent.
number of CaCl₂ moles in 500 g water - 0.045 mol
Therefore number of moles in 1 kg water - 0.045 mol / 500g x 1000 g = 0.090 mol
Molality of CaCl₂ - 0.090 mol/kg

5 0

4 years ago