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jeka57 [31]
3 years ago
7

On the moon the acceleration due to gravity is 1.6 ms if an object has a mass on earth of 5.1kg what is its mass on the moon

Chemistry
1 answer:
Y_Kistochka [10]3 years ago
3 0

Answer:

Mass on the moon is 5.1 kg.

Explanation:

Given mass on earth = 5.1 kg

acceleration due to gravity on moon = 1.6 ms⁻².

The mass on the moon does not change because mass is the quantity of matter in the body regardless of its volume or  of any forces acting on it.

Whereas the weight of the object changed.

Weight is the the force exerted on a body which is related to mass and expressed as W = mg.

W = weight of object   m = mass of object      and g = gravitational force

So\ in\ this\ case\ W\ =1.6\times\ 5.1\ =8.16N

N stands for newton unit of weight.

Thus mass of the object on moon is same as on earth which is 5.1 kg.

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Dinitrogen tetraoxide is a colorless gas at room temperature. It can dissociate into nitrogen dioxide, which is a reddish brown
Alenkinab [10]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The calculation and filling of the table based on the ratio indicated by the column is shown on the second uploaded image

1 This is True

2 This is False

3 This is False

4 This is  True

5 This is  True

Explanation:

Based on your calculations, indicate whether each statement is True (T) or False (F):

Considering the first statement

Looking at the question we can see that the first statement is TRUE this because  all the three experiment have different initial concentration (3.5,3.0 & 2.6)

Considering the second  statement

 Looking at table we can see that the second statement is FALSE  because  

the ratio value of [NO2]/ [N2O4] in all three experiment have different value  1.4 , 1.5 & 1.7 is not a constant value.

Considering the Third  statement

Looking at the table we can see that the third statement is FALSE this  is because the ratio value of 2[NO2]/ [N2O4] in all three experiment have different value 2.7, 3.0 & 3.3 is not a constant value.

Considering the Fourth  statement

Looking at table we can see that the fourth statement is TRUE this is because the ratio value of[NO2]^{2}/ [N2O4] in all three experiment have same value 3.9, 3.9 & 3.9 is a constant value.

Considering the Fifth  statement

Looking at table we can see that the fifth statement is TRUE this is because

each having different set of Equilibrium concentration

3 0
4 years ago
Which of the following UV rays has the highest energy that damages your skin the most?
Sergio [31]
The answer is b) UVB
7 0
2 years ago
50 POINTS!
Varvara68 [4.7K]
5.5/-38 / 5.5 = 1/-6.9 x 1.3 = 1.3/-8.98
unit volume/temperature x searching amount

i’d say the temperature would be -8.98 C simply - I don’t know what formula youd use for this

6 0
3 years ago
From the data table given, calculate the missing quantity
ololo11 [35]

<u>Answer:</u> The initial volume of the gas is 16.55 mL

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=?mL\\T_1=300K\\V_2=16mL\\T_2=290K

Putting values in above equation, we get:

\frac{V_1}{300K}=\frac{16mL}{290K}\\\\V_1=\frac{16\times 300}{290}=16.55mL

Hence, the initial volume of the gas is 16.55 mL

4 0
4 years ago
What is the enthalpy for the following reaction? overall: C + 1/2O2→CO
telo118 [61]

Answer:

ΔH°(f) = -110.5 Kj/mole (exothermic)

Explanation:

C + 1/2O₂ => CO

This is asking for the 'Standard Heat of Formation (ΔH°(f)* for carbon monoxide (CO). Values for many compounds can be found in the appendix of most college general chemistry text books. From Ebbing & Gammon, 11th edition, General Chemistry, Appendix C, page 8A.

*Standard Heat of Formation by definition is the heat gained or lost on formation of a substance (compound) from its basic elements in standard state.

The ΔH°(f) values as indicated are found in the appendix of most college chemistry texts. By choosing any compound, one can determine the standard heat of formation equation for the substance of interest. For example, consider Magnesium Carbonate; MgCO₃(s).The basic standard states of each element is found in the Appendix on Thermodynamic Properties for Substances at 25°C & 1 atm. having  ΔH°(f) values = 0.00 Kj/mole. All elements in standard state have a 0 Kj/mol. See appendix and note that under the ΔH°(f) symbol some substances have 0.00 Kj/mol values. The associated element will be in basic standard state,

Standard Heat of Formation Equation for formation of Magnesium Carbonate;

Mg°(s) + C°(gpt)* + 3/2O₂(g) => MgCO₃(s) ; ΔH°(f) = -1111.7 Kj/mole

* gpt => graphite

4 0
3 years ago
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