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Helen [10]
2 years ago
8

An object accelerates 12.0 m/s2 when a force of 6.0 newtons is applied to it. What is the mass of the object?

Chemistry
1 answer:
dedylja [7]2 years ago
5 0

Answer:

The mass of object is 0.5 Kg.

Explanation:

Given data:

Acceleration of object = 12.0 m/s²

Force on object = 6.0 N

Mass of object = ?

Solution:

Formula:

F = m×a

F = force

m = mass

a = acceleration

Now we will put the values in formula.

6.0 N = m  × 12.0 m/s²

m =  6.0 N / 12.0 m/s²  

   ( N = kg.m/s²)

m = 0.5 kg

The mass of object is 0.5 Kg.

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A small cube of iron and a large flat sheet of iron contain the same volume. Which one will completely rust first? Explain why.
allochka39001 [22]
Flat as more oxygen and water can react over it think of it like this would a cube rust faster than a sheet
6 0
1 year ago
You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain
IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of O_2:-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{49.8\ g}{32\ g/mol}

Moles_{O_2}= 1.55625\ mol

Since pressure and volume are constant, we can use the Avogadro's law  as:-

\frac {V_1}{n_1}=\frac {V_2}{n_2}

Given ,  

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

\frac {V_1}{n_1}=\frac {V_2}{n_2}

\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of CO_2 = 44.0 g/mol

Mass of CO_2 = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

5 0
1 year ago
What is the stoichiometric ratio between BaCl2 and NaCl
bixtya [17]
<span>BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
7 0
2 years ago
For a pure substance, the liquid and gaseous phases can only coexist for a single value of the pressure at a given temperature.
anastassius [24]

Answer:

No, it is not.

Explanation:

Most solutions do not behave ideally. Designating two volatile  substances as A and B, we can consider the following two cases:

Case 1: If the intermolecular forces between A and B molecules are weaker than  those between A molecules and between B molecules, then there is a greater tendency  for these molecules to leave the solution than in the case of an ideal solution. Consequently,  the vapor pressure of the solution is greater than the sum of the vapor  pressures as predicted by Raoult’s law for the same concentration. This behavior gives  rise to the positive deviation.

Case 2: If A molecules attract B molecules more strongly than they do their own  kind, the vapor pressure of the solution is less than the sum of the vapor pressures as  predicted by Raoult’s law. Here we have a negative deviation.

The benzene/toluene system is an exception, since that solution behaves ideally.

8 0
2 years ago
After a certain pesticide compound is applied to crops, its decomposition is a first-order reaction with a half-life of 56 days.
Anna35 [415]

The rate constant, k, for the decomposition reaction :  k = 0.0124 / days

<h3>Further explanation</h3>

Given

The half-life of 56 days

Required

The rate constant, k

Solution

For first-order, rate law : ln[A]=−kt+ln[A]o

The half-life :  the time required to reduce to half of its initial value.

The half life :

t1/2 = (ln 2) / k

k = (ln 2) / t1/2

k = 0.693 / 56 days

k = 0.0124 / days

8 0
2 years ago
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