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2 years ago

If a force of 1.4N is applied to a block with a mass of 7kg, what is the acceleration of the block in meters per second squared?

Chemistry

2 answers:

Troyanec [42]2 years ago

8 0

8,394m2 for 7kg with force 1.4

Anton [14]2 years ago

5 0

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{1.4}{7} = \frac{1}{5} \\$

We have the final answer as

Hope this helps you

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The chemical equation below is correctly balanced.

DIA [1.3K]

Answer:

44.8 L of O2 will react (option D)

Explanation:

Step 1: Data given

Number of moles of SO2 = 4.00 moles

STP = Pressure = 1 atm and temperature = 273 K

Step 2: The balanced equation

2 SO2(g) + O2(g) → 2 SO3(g)

Step 3: Calculate moles of O2

For 2 moles SO2, we need 1 mol O2 to produce 2 moles SO3

For 4.00 moles SO2 we need 4.00 / 2 = 2.00 moles O2

Step 4: Calculate volume of O2

For 1 mol we have a volume of 22.4 L

V = (n*R*T)/ p

V = (2.00 * 0.08206 * 273)/p

V = 44.8 L

For 2.00 moles we have a volume of 2*22.4 = 44.8 L

44.8 L of O2 will react (option D)

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3 years ago

Geraldo wants to model the effect of acids on the pH of

Furkat [3]

Answer: Pesticide and less than 6.65

Explanation:

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3 years ago

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Sphinxa [80]

Answer:

Condensation and Depositon

Explanation:

Condensation is from gas to liquid

Deposition is from gas to solid

7 0

2 years ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

<u>The activation energy for this reaction = 23 kJ/mol.</u>

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2 years ago

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sattari [20]

Ngl idk but just be a nice person and mark as brainliest please

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3 years ago

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