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2 years ago

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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Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A

lesantik [10]

<u>Answer:</u> The chemical equations and equilibrium constant expression for each ionization steps is written below.

<u>Explanation:</u>

The chemical formula of carbonic acid is $H_2CO_3$ . It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water

The chemical equation for the first dissociation of carbonic acid follows:

$H_2CO_3(aq.)\rightleftharpoons H^+(aq.)+HCO_3^-(aq.)$

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}$

The chemical equation for the second dissociation of carbonic acid follows:

$HCO_3^-(aq.)\rightarrow H^+(aq.)+CO_3^{2-}(aq.)$

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}$

Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.

6 0

2 years ago

Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,

8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0

2 years ago

For the reaction 2N2O5(g) <---> 4NO2(g) + O2(g), the following data were colected:

KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0

2 years ago

What is water's density at 93 ∘C? Assume a constant coefficient of volume expansion. Express your answer with the appropriate un

Ganezh [65]

Answer:

982.5 kg/m³

Explanation:

When the temperature of a fluid increases, it dilates, and because of the variation of the volume, it's density will vary too. The density can be calculated by the expression:

ρ₁ = ρ₀/(1 + β*(t₁ - t₀))

Where ρ₁ is the final density, ρ₀ the initial density, β is the constant coefficient of volume expansion, t₁ the final temperature, and t₀ the initial temperature.

At t₀ = 4°C, the water desity is ρ₀ = 1,000 kg/m³. The value of the constant for water is β = 0.0002 m³/m³ °C, so, for t₁ = 93°C

ρ₁ = 1,000/(1 + 0.0002*(93 - 4))

ρ₁ = 1,000/(1+ 0.0178)

ρ₁ = 982.5 kg/m³

3 0

2 years ago

The shape of an atomic orbital is associated with

Diano4ka-milaya [45]

The properties of the atomic orbital are actually dependent on the quantum numbers.

size of atomic orbital: governed by the principal quantum number (n)

shape of atomic orbital: governed by the angular momentum quantum number (l)

orientation in space: governed by the magnetic quantum number (ml)

Since we are asked about the shape, hence the correct answer is:

angular momentum quantum number (l)

8 0

2 years ago