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lesya [120]
3 years ago
12

Explain in detail what happens when an acid and a base mix. Try to use the words ions, attraction, neutral and bonding in your a

nswer.
( I understand how to use all keywords except from attraction, when does attraction occur in neutralization?)
Chemistry
1 answer:
Anton [14]3 years ago
5 0

Answer:

when acid and the base mix it is bonding together and that makes it have a little blast but because of the acid

Explanation:

acid is the mane thing that makes 50% of things blast like as if i were to put baking soada and acid it would blast.  

Click the heart if this was helpful

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Add 92.2 km to 9426 m and report the answer in
Mashcka [7]

to convert m into km we have to divide it by 1000.so,

9426/1000=9.426km

92.2km+9.426km=101.626km.

3 0
3 years ago
"A crosslinked copolymer consists of 57 wt% ethylene (C2H4) repeat units and 43 wt% propylene (C3H6) repeat units. Determine the
ch4aika [34]

Explanation:

Percentage ethylene by weight = 57%

Percentage propylene by weight = 43%

Suppose in 100 grams of polymer:

Mass of ethylene = 57 g

Mass of propylene = 43 g

Moles of ethylene = \frac{57 g}{28 g/mol}=2.036 mol

Moles of propylene = \frac{43g}{42g/mol}=1.024 mol

1 mole = N_A =6.022\times 10^{23} molecules/ atoms

Units of ethylene = 2.036 mol\times N_A

Units of propylene = 1.024 mol\times N_A

a) Fraction of ethylene units:

=\frac{2.036 mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{509}{765}

b ) Fraction of propylene units:

=\frac{1.024mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{256}{765}

3 0
3 years ago
Plz answer thx<br><br><br> ​How many moles are in 2.5 x 10^18 grams of water?
alex41 [277]

Answer:

1.39\times 10^{17} moles of water in 2.5\times 10^{18} g of water.

Explanation:

Mass of water = m = 2.5\times 10^{18} g

Molar mass of water = M = 18 g/mol

Moles = n = \frac{m}{M}

n=\frac{2.5\times 10^{18} g}{18 g/mol}=1.39\times 10^{17} moles

So, there are 1.39\times 10^{17} moles of water in 2.5\times 10^{18} g of water.

3 0
3 years ago
What volume of water is required to prepare 0.1 M H3PO4 from 100 ml of 0.5 M solution?
ExtremeBDS [4]

Answer: A volume of 500 mL water is required to prepare 0.1 M H_{3}PO_{4} from 100 ml of 0.5 M solution.

Explanation:

Given: M_{1} = 0.1 M,    V_{1} = ?

M_{2} = 0.5 M,       V_{2} = 100 mL

Formula used to calculate the volume of water is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.1 M \times V_{1} = 0.5 M \times 100 mL\\V_{1} = 500 mL

Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M H_{3}PO_{4} from 100 ml of 0.5 M solution.

7 0
3 years ago
The value of the rate constant for a gas phase reaction can be changed by increasing the A. temperature of the reaction vessel.
prohojiy [21]

Answer:

temperature of the reaction vessel

Explanation:

temperature of the reaction vessel

3 0
2 years ago
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