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beks73 [17]
4 years ago
14

True or false: scientists make observations very carefully science

Chemistry
2 answers:
posledela4 years ago
8 0
True hope this helped u
Olenka [21]4 years ago
3 0
True, scientists do make observations very carefully.
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In stoichiometric calculations, which quantity MUST be used to convert from one chemical substance to another?
Cloud [144]

The answer is

C) mole ratio

4 0
2 years ago
In a titration, 4.7 g of an acid (HX) requires 32.6 mL of 0.54 M NaOH(aq) for complete reaction. What is the molar mass of the a
katrin2010 [14]

Answer : The molar mass of an acid is 266.985 g/mole

Explanation : Given,

Mass of an acid (HX) = 4.7 g

Volume of NaOH = 32.6 ml = 0.0326 L

Molarity of NaOH = 0.54 M = 0.54 mole/L

First we have to calculate the moles of NaOH.

\text{Moles of }NaOH=\text{Molarity of }NaOH\times \text{Volume of solution}=0.54mole/L\times 0.0326L=0.017604mole

Now we have to calculate the moles of an acid.

In the titration, the moles of an acid will be equal to the moles of NaOH.

Moles of an acid = Moles of NaOH = 0.017604 mole

Now we have to calculate the molar mass of and acid.

\text{Moles of an acid}=\frac{\text{Mass of an acid}}{\text{Molar mass of an acid}}

Now put all the given values in this formula, we get:

0.017604mole=\frac{4.7g}{\text{Molar mass of an acid}}

\text{Molar mass of an acid}=266.985g/mole

Therefore, the molar mass of an acid is 266.985 g/mole

3 0
3 years ago
All atoms of a given element are identical was proven wrong because of the existence or
vagabundo [1.1K]

Answer:

they are all similar

Explanation:

6 0
3 years ago
Which statement summarizes the difference between mass and weight?
alexgriva [62]

Answer:

Mass is the amount of matter in an object.

Weight is how much an object weighs.

Hope this helps!

6 0
3 years ago
How many milliliters of 0.0896M LiOH are required to titrate 25.0 mL of 0.0759M HBr to the equivalence point?
slavikrds [6]

Answer:

V_{LiOH}=21.8mL

Explanation:

Hello,

In this case, during titration at the equivalence point, we find that the moles of the base equals the moles of the acid:

n_{LiOH}=n_{HBr}

That it terms of molarities and volumes we have:

M_{LiOH}V_{LiOH}=M_{HBr}V_{HBr}

Next, solving for the volume of lithium hydroxide we obtain:

V_{LiOH}=\frac{M_{HBr}V_{HBr}}{M_{LiOH}} =\frac{0.0759M*25.0mL}{0.0896M} \\\\V_{LiOH}=21.8mL

Best regards.

8 0
3 years ago
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