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kap26 [50]
3 years ago
14

#5 6th grade science

Chemistry
1 answer:
Alex777 [14]3 years ago
6 0

Answer:

I'm pretty sure it's D.

Explanation:

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The balanced equation for the combustion of Hexane,C6H14 , is
shusha [124]

Answer:

2C _{6}H _{14} +  19O _{2}→12CO _{2} +14H _{2} O

3 0
3 years ago
Help me pls I put 49 point( every single point that I have) pls help me
Aliun [14]

Answer:

Chemical reaction, compound, molecule, covalent bonds, ionic bonds, ions. (first page)

Oxygen and hydrogen, 1 hydrogen 2 oxygen, H2O, a covalent bond, then mark off is always a liquid. (Second page)

Explanation:

6 0
3 years ago
What volume of stock solution and water must you add to prepare 36.25ml of a 1.25M solution
Serjik [45]

Given :

Number of moles , n = 36.25 mol .

Molarity , M = 1.25 M .

To Find :

The volume of water required .

Solution :

Moarity is given by :

M=\dfrac{n}{V}

So , V=\dfrac{n}{M}

Here , n is number of moles and M is molarity .

Putting all values in above equation , we get :

V=\dfrac{36.25}{1.25}\\\\V=29\ L

Therefore , volume of water required is 29 L .

5 0
3 years ago
A volume of 30.0 mL of a 0.140 M HNO3 solution is titrated with 0.570 M KOH. Calculate the volume of KOH required to reach the e
igomit [66]

Answer:

7.37 mL of KOH

Explanation:

So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,

HNO3 + KOH → KNO3 + H2O

Step 1 : The moles of HNO3 here can be calculated through the given molar mass (  0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.

Mol of NHO3 = 0.140 M * 30 / 1000 L = 0.140 M * 0.03 L = .0042 mol

Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,

0.0042 mol HNO2 * ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH

From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,

Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).

Volume of KOH = 0.0042 mol * ( 1 L / 0.570 mol ) * ( 1000 mL / 1 L ) = 7.37 mL of KOH

5 0
4 years ago
How many grams of H2 can be formed from 54.6 grams of NH3 in the following reaction? 2NH3 (g) --> 3H2 (g) + N2 (g)
Strike441 [17]

Answer:

9.64g

Explanation:

The balanced equation for the reaction is given below:

2NH3 (g) —> 3H2 (g) + N2 (g)

Next, we need to calculate the mass NH3 that decomposed and the mass of H2 produced from the balanced equation. This is illustrated below:

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

Mass of NH3 that decomposed from the balanced equation = 2 x 17 = 34g

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 produced from the balanced equation = 3 x 2 = 6g.

Now, we can obtain the mass of H2 formed from 54.6g of NH3 as follow:

From the balanced equation above,

34g of NH3 decomposed to produce 6g of H2.

Therefore, 54.6g of NH3 will decompose to produce = (54.6x6)/34 = 9.64g of H2

Therefore, 9.64g of H2 can be obtained from 54.6g of NH3.

3 0
3 years ago
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