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Mama L [17]
3 years ago
9

If you throw a ball into the air, the force of ________ makes the ball fall back toward Earth.

Chemistry
2 answers:
larisa86 [58]3 years ago
7 0
B.) gravity
Explanation: gravity makes sense
Tanya [424]3 years ago
6 0
Acceleration is when something is being moved forward or back, motion is just movement, friction is two or more things rubbing together, so the answer should be B. gravity
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Nitrogen forms more oxides than any other element. The percents by mass of in three different nitrogen oxides are (1) (II) and (
Tema [17]

Complete question;

Nitrogen forms more oxides than any other element. The percents by mass of N in three different nitrogen oxides are (|) 46.69%;(II) 36.85 %; (III) 25.94%. For each compound, determine (a) the simplest whole-number ratio of N to O, and (b) the number of grams of oxygen per 1.00 g of nitrogen.

Answer:

a. (i) The ratio is 1:1 , the formula = NO  (ii)The ratio is 1 : 1.5 which is  2 : 3, the formula = N₂O₃  (iii) The ratio is 1 : 2.5 which is 2:5 , the formula = N₂O₅

b. (i)number of grams of oxygen = 53.31/46.69 = 1.14 g

(ii)number of grams of oxygen = 63.15/36.8 = 1.71 g

(iii)number of grams of oxygen = 74.06/25.94 = 2.855 g

Explanation:

a.

(i) The percentage by mass of the nitrogen in Nitrogen oxide (i) is 46.69% which is taken as 46.69 grams . Since the other element is oxygen the mass of oxygen will be 100 - 46.69 = 53.31 grams.

The relative atomic mass of Nitrogen and oxygen is 14 amu and 16 amu respectively.

Therefore, to know the whole number ratio of N and O we find the number of moles.

number of moles of N = 46.69/14 = 3.335

number of moles of O = 53.31/16 = 3.332

The ratio is 1:1 , the formula = NO

(ii)

number of moles of N = 36.85/14 = 2.632

number of moles of O = 63.15/16 = 3.947

The ratio is 1 : 1.5 which is  2 : 3, the formula = N₂O₃

(iii)

number of moles of N = 25.94/14 = 1.85

number of moles of O = 74.06/16 = 4.63

The ratio is 1 : 2.5 which is 2:5 , the formula = N₂O₅

b.

(i) 46.69 g of nitrogen  = 53.31 g of oxygen

1 g of nitrogen = ? of Oxygen

number of grams of oxygen = 53.31/46.69 = 1.14 g

(ii)

Using similar method in b(i)

number of grams of oxygen = 63.15/36.8 = 1.71 g

(iii)

Using similar method in b(i)

number of grams of oxygen = 74.06/25.94 = 2.855 g

3 0
3 years ago
State relationship between the level of dissolved oxygen and water temperature
Artemon [7]
More dissolved oxygen is present in water with a lower temperature compared to water with a higher temperature. The reason for this inverse relationship between dissolved oxygen and temperature is that the solubility of a gas in a liquid is an equilibrium phenomenon.
8 0
4 years ago
You transfer a sample of a gas at 17°C from a volume of 5.67 L and 1.10 atm to a container at 37°C that has a pressure of 1.10 a
laila [671]

Explanation:

V1T1 = V2T2

V2=V1T1/T2

= 5.67*17/37= 2.6

5 0
4 years ago
The chemical formula for magnesium oxide is MgO. A chemist determined by measurements that 0.030 moles of magnesium oxide partic
tamaranim1 [39]

Answer:

1.209g of MgO participates

Explanation:

In this problem, we have 0.030 moles of MgO that participates in a particular reaction.

And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.

To convert moles to grams we need to use molar mass of the compound:

<em>1 atom of Mg has a molar mass of 24.3g/mol</em>

<em>1 atom of O has a molar mass of 16g/mol</em>

<em />

That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol

And mass of 0.030 moles of MgO is:

0.030 moles MgO * (40.3g/mol) =

<h3>1.209g of MgO participates</h3>
3 0
3 years ago
What mass of oxygen is needed to burn 54.0 grams of butane c4h10
Jobisdone [24]

The moles of oxygen required to burn Butane is 6 moles.

<h3>What is a Combustion Reaction?</h3>

A reaction in which fuel gets oxidised by an oxidising agent producing a large amount of heat is called a combustion reaction.

In this question

Butane is burnt with oxygen

Molar mass of C₄H₁₀ = (12.0×4 + 1.0×10) g/mol = 58.0 g/mol

Molar mass of O₂ = 16.0×2 g/mol = 32.0 g/mol

Balanced equation for the reaction:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Mole ratio C₄H₁₀ : O₂ = 2 : 13

The given mass = 54grams

moles = 54/58 = 0.93 moles

The mole of oxygen required =

0.93/ x = 2/13

0.93*13/2 = x

x = 6.045 moles

Therefore 6 moles of oxygen are required to burn Butane.

To know more about Combustion Reaction

brainly.com/question/12172040

#SPJ1

6 0
2 years ago
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