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Svetlanka [38]
3 years ago
8

What should be ph-value to be strong acid?

Chemistry
2 answers:
irakobra [83]3 years ago
6 0

<em>what should be ph-value to be strong acid?</em>

<em>what should be ph-value to be strong acid?</em><em>c</em><em>.</em><em> Less than </em><em>2</em><em>.</em><em>.</em><em>.</em><em>.</em>

alina1380 [7]3 years ago
6 0

Answer:

the answer is option a because strong acids are less than 4

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when washing soda is mixed with lemon juice, bubbles are formed with the evolution of gas. what type of change is it. ​
aleksandr82 [10.1K]

Answer:

it is a chemical change

Explanation:

it is a chemical change since the lemon juice can not be returned to its original form some part of it had been converted to gas form

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If a drop of blood is 0.05 mL, how many drops of blood are in the blood collection tube that holds 2 mL?
Irina18 [472]
40 drops of blood in a tube that holds 2 mL
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Predict the missing product of this equation<br><br><br>1 MgF2 + 1 Li2CO3 -&gt; 1 ______ +2LiF
ch4aika [34]

Answer:

MgCO₃

Explanation:

From the question given above, we obtained:

MgF₂ + Li₂CO₃ —> __ + 2LiF

The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:

MgF₂ (aq) —> Mg²⁺ + 2F¯

Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯

MgF₂ + Li₂CO₃ —>

Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:

MgF₂ + Li₂CO₃ —> __ + 2LiF

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Therefore, the missing part of the equation is MgCO₃

8 0
3 years ago
Demonstrate an understanding of Stoichiometry by describing and calculating the components required for a stoichiometric evaluat
NeTakaya

Explanation:

Iam sorry I don't know but why Iam messaging iss because when more people message it usually appears to more people so someone else will be able to help you:)

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Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO
tekilochka [14]

Answer: There are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

So, the number of atoms present in 0.67 moles are as follows.

0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms

Thus, we can conclude that there are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

7 0
3 years ago
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