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Xelga [282]
3 years ago
11

A(n) ______ Is a group of organs that work together to perform a com

Chemistry
1 answer:
soldier1979 [14.2K]3 years ago
5 0

Answer:

an organ system  Is a group of organs that work together to perform a com

Explanation:

that is the answer organ system

now plz give me 5 stars and press on the heart plz

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I will not stop asking until someone answer hurry up already
Reil [10]

Answer:

Gravity.

Explanation:

4 0
3 years ago
A reaction between ammonia gas and chlorine gas produces
7nadin3 [17]

Answer:

32.25

Explanation:

8NH3 + 3Cl2 → 6NH4Cl + N2

NH3 = 17 g/mol

number of moles = 1

NH4Cl= 43 g/mol

number of moles = 3/4

mass= 43 × ¾ ≈ 32.25 g

7 0
3 years ago
Read 2 more answers
Express 6009m in scientific notation
Andre45 [30]
6.009 x 10^3. You have to put 6009 into a decimal that is less than 10. Count backwards from 9 until you get to a single digit. You would move the decimal back three places making it 10^3
4 0
3 years ago
Read 2 more answers
1. Take the reaction: NH3 + O2 + NO + H2O. In an experiment, 3.25g of NH3 are allowed
Rudiy27

Answer:

5.74g of NO

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

4NH3 + 5O2 —> 4NO + 6H2O

Step 2:

Determination of the masses of NH3 and O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar Mass of NO = 14 + 16 = 30g/mol

Mass of NO from the balanced equation = 4 x 30 = 120g

From the balanced equation above,

68g of NH3 reacted with 160g of O2 to produce 120g of NO.

Step 3:

Determination of the limiting reactant.

We need to determine the limiting because it will be used to calculate the maximum yield of the reaction. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Therefore, 3.25g of NH3 will react with = (3.25 x 160)/68 = 7.65g of O2.

From the simple illustration above, we can see that lesser mass of O2 is needed to react with 3.25g of NH3. Therefore, NH3 is the limiting reactant while O2 is the excess reactant.

Step 4:

Determination of the mass of NO produced from the reaction.

In this case the limiting reactant will be used because all of it were used in the reaction.

The limiting reactant is NH3.

From the balanced equation above,

68g of NH3 reacted to produce 120g of NO.

Therefore, 3.25g of NH3 will react to produce = (3.25 x 120)/68 = 5.74g of NO.

From the calculations made above, 5.74g of NO is produced.

4 0
3 years ago
How many hydrogen atoms are in an acyclic alkane with 8 carbon atoms?
vfiekz [6]
The general formula of acyclic alkane is CnH2n+2. 

<span>When n = 8 : </span>
<span>Number of hydrogen atoms = 2n + 2 = 18</span>
7 0
3 years ago
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