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2 years ago

What is 24.29 g + 4.336 g - 17.900 g

Chemistry

2 answers:

Jet001 [13]2 years ago

8 0

10.726g or 10.73g if rounded to hundredths place

choli [55]2 years ago

5 0

Answer:

10.726g

Explanation:

You might be interested in

How many grams of sodium are needed to react completely with 19

gtnhenbr [62]

Answer:

46 g of sodium

Explanation:

Sodium reacts vigorously with fluoride gas to form NaF as shown

2Na (s) + F2 (g) ------> 2NaF (s) (Na = 23, F = 19)

2 moles of Na reacts with 1 mole of F2 to produce NaF

By calculating the molar mass of the elements involved in the reaction then multiplied by the mole, the mass can be obtained.

23 * 2 g/mol of Na reacts with 1 * 19 g/mol of F2

46 g/mol of Na reacts with 19 g/mol of F2 to produce NaF

Since the mole ratio is 2 to 1 and 19 g of F2 is used for the reaction, 46 g of sodium will be consumed for the reaction to be achieved.

6 0

3 years ago

Explain how an element that does not react easily with other elements can be useful.

BaLLatris [955]

Elements that are unreactive can be useful for 'isolating' other volatile elements. For example, an element may be highly reactive with oxygen, so keeping it in air is not a safe option. In such cases, it may be useful to have a noble gas to store the element safely and ensure its stability. It may also have a purpose in experiments where it may act as a controlling agent or a container for the other reactions.

5 0

3 years ago

2al2o3 yields to 4al + 3o2 .. how many miles of oxygen are produced from the decomposition of 1.26 mil of Al2O3?

gladu [14]

Answer:

$\large \boxed{\text{2.52 mol Al}}$

Explanation:

2Al₂O₃ ⟶ 4Al +3O₂

n/mol: 1.26

The molar ratio is 4 mol Al:2 mol Al₂O₃.

$\text{Moles of Al} = \text{1.26 mol Al$_{2}$O}_{3} \times \dfrac{\text{4 mol Al}}{\text{2 mol Al$_{2}$O}_{3}}= \textbf{2.52 mol Al}\\\\\text{The reaction produces $\large \boxed{\textbf{2.52 mol Al}}$}$

7 0

3 years ago

In a combustion chamber, ethane (c2h6) is burned at a rate of 9 kg/h with air that enters the combustion chamber at a rate of 17

nekit [7.7K]

The percentage of excess air used during combustion process of ethane will be 37 %.

Burning, also known as combustion, would be a high-temperature highly exothermic chemical process that occurs when an oxidant, typically atmospheric oxygen, interacts with a fuel to generate oxidized, frequently gaseous products in a mixture known as smoke.

Calculation of percentage of air .

$C_{2} H_{6} + (1-x)+a(O_{2} +3.76N_{2} )=bCO_{2} + cH_{2} O + axO_{2} + 3.76dN_{2} .$

Mair=Mair/Rin

$( MN)O_{2}$ + $(MN)N_{2}$ ÷ $(MN)O_{2}$ + $(MN)N_{2}$ + $(MN)C_{2} H_{6} .$

33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1

= 176/176+8

X= 0.37

0.37 × 100

X= 37%

Therefore, the percentage of excess air used during combustion process of ethane will be 37 %.

To know more about combustion process

brainly.com/question/13153771

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3 0

1 year ago

A helium-filled balloon has a volume of 2.75 L at 20.°C. The volume of the

omeli [17]

Answer: The outside temperature is 262.1K or -10.9°C

Explanation:

Given that,

Original volume V1 = 2.75 L

Original temperature T1 = 20.0°C

Since unit of temperature is Kelvin, convert 20°C to Kelvin by adding 273 to 20°C

(20.0°C + 273K = 293K)

Outside volume V2 = 2.46 L

Outside temperature T2 = ?

Since volume and temperature are given while pressure is constant, apply the formula for Charle's law

V1/T1 = V2/T2

2.75L/293K = 2.46L/T2

To get the value of T2, cross multiply

2.75L x T2 = 2.46L x 293K

2.75LT2 = 720.78LK

Divide both sides by 2.75L

2.75LT2/ 2.75L = 720.78LK/2.75L

T2 = 262.1K

Hence, the outside temperature in Kelvin is 262.1K. Then, obtain outside temperature in Celsius by subtracting 273 from 262.1K

i.e 262.1K - 273 = -10.9°C

Thus, the outside temperature is 262.1K, or -10.9°C

5 0

3 years ago