All categories

3 years ago

A force of 20 N acts upon 5 kg block caculate the accerleration of the object

Chemistry

1 answer:

Vedmedyk [2.9K]3 years ago

4 0

Answer:

$\boxed{\text{4 m $\cdot$ s$^{-2}$}}$

Explanation:

F = ma

$a = \dfrac{F}{m} = \dfrac{\text{20 N}}{\text{5 kg}} \times \dfrac{\text{1 kg$\cdot$ m $\cdot$ s$^{-2}$}}{\text{1 N}} = \text{4 m $\cdot$ s$^{-2}$}\\\\\text{The acceleration of the object is } \boxed{\textbf{4 m $\cdot$ s$^{\mathbf{-2}}$}}$

You might be interested in

-5 - -2 plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz helppppppppppppppppppppppppppppppp meee

Zepler [3.9K]

The answer would be -3

4 0

3 years ago

Read 2 more answers

Helpppp me please please

kipiarov [429]

The answer is letter d

3 0

3 years ago

Read 2 more answers

What might be a possible outcome if you left a soda in a glass bottle in the freezer for a very long time?

Scrat [10]

A the soda will increase in volume

6 0

2 years ago

2) 2KClO3 --> 2KCl + 3O2

aleksandrvk [35]

$2 \text{ KClO}_3 \to 2 \text{ KCl}+3\text{ O}_2$

$2 \text{ mols of KClO}_3 \equiv 3 \text{ mols of O}_2$

$19 \text{ mols of KClO}_3 \equiv 3\cdot 9,5 \text{ mols of O}_2$

$\boxed{19 \text{ mols of KClO}_3 \equiv 28,5 \text{ mols of O}_2}$

$2 \text{ mols of KClO}_3 \equiv 2 \text{ mols of KCl}$

$62 \text{ mol of KClO}_3 \equiv 62 \text{ mol of KCl}$

Using the atomic mass given in the periodic table:

$62\cdot(39+35,5+16\cdot3) \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}$

$62\cdot122,5 \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}$

$7595 \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}$

$\boxed{7,595 \text{ kg of KClO}_3 \equiv 62 \text{ mol of KCl}}$

$2 \text{ KCl}+3\text{ O}_2\to 2 \text{ KClO}_3$

$3 \text{ mols of O}_2 \equiv 2 \text{ mols of KCl}$

Using the atomic mass given in the periodic table:

$3\cdot(2\cdot 16) \text{ g of O}_2 \equiv 2\cdot(39+35,5) \text{ g of KCl}$

$96\text{ g of O}_2 \equiv 149\text{ g of KCl}$

$\dfrac{39}{149}\cdot 96\text{ g of O}_2 \equiv \dfrac{39}{149}\cdot 149\text{ g of KCl}$

$\boxed{25,13\text{ g of O}_2 \equiv 39\text{ g of KCl}}$

This result is an aproximation.

8 0

3 years ago

Be sure to answer all parts. one of the most important industrial sources of ethanol is the reaction of steam with ethene derive

lions [1.4K]

Answer: 2.17x10⁻³ atm

Explanation:

First, we must write the balanced chemical equation for the process:

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)

The chemical reactions that occur in a closed container can reach a state of chemical equilibrium that is characterized because the concentrations of the reactants and products remain constant over time. The equilibrium constant of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

The equilibrium constant (K) is expressed as the ratio between the molar concentrations (mol/L) of reactants and products. Its value in a chemical reaction depends on the temperature, so it must always be specified.

We will use the the equilibrium constant Kc of the reaction to calculate partial pressure of ethene. The constant Kc for the above reaction is,

Kc = $\frac{[C_{2} H_{5}OH]}{[H_{2}O][C_{2} H_{4}]}$

According to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the gas constant (0.082057 atm L / mol K) .

We can use the ideal gas law to determine the molar concentrations ([x] = n / V) from the gas pressures of ethanol and water, assuming that all gases involved behave as ideal gases. In this way,

PV = nRT → P = (n/V) RT → P = [x] RT → [x] = P / RT

So,

$[C_{2} H_{5}OH] = \frac{200 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 4.06 \frac{mol}{L}$

$[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}$

So, the molar concentration of ethene (C₂H₄) will be,

$[C_{2} H_{4}] = \frac{[C_{2} H_{5}OH]}{[H_{2}O] x Kc} = \frac{4.06 \frac{mol}{L} }{8.12 \frac{mol}{L}x9.00 x 10^{3} \frac{L}{mol} } = 5.56 x 10^{-5}\frac{mol}{L}$

Then, according to the law of ideal gases,

$P_{C_{2} H_{4}} = [C_{2} H_{4}]RT = 5.56 x 10^{-5} \frac{mol}{L} x 0.082057 \frac{atm L}{mol K} x 600 K = 2.17x10^{-3} atm$

So, when the partial pressure of ethanol is 200 atm and the partial pressure of water is 400 atm, the partial pressure of ethene at 600 K is 2.17x10⁻³ atm.

7 0

3 years ago