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steposvetlana [31]
3 years ago
13

How many milliliters are equal to 0.399195L

Chemistry
1 answer:
Gennadij [26K]3 years ago
5 0
0.399195 L equals 399.195 mL I believe
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How much volume would a 834.01g pile of sugar have given that it has a density of 1.59g/mL?
boyakko [2]

Answer:

v = 534.5mL

m = 597.15g

Density = 9.23g/mL

Density = 9.125g/mL

Explanation:

Density = mass/ volume

For the first question

Density = 1.59g/mL

Mass = 834.01g

Volume = ?

Using the above formula we have 1.59 = 834.01/v

v = 834.01/1.59

v = 534.5mL

For the second question

Density =0.9167g/mL

Volume = 651.41mL

Mass =?

Using the above formula we have

0.9167 =m/651.41

Cross multiply

m = 0.9167 x 651.41

m = 597.15g

For the third question

Mass =803.44g

Volume=87.03mL

Density =?

Density = 803.44/87.03

= 9.23g/mL

For the fourth

Density = 56.85/6.23

= 9.125g/mL

7 0
3 years ago
What do cubic measurements, live cm3, mesure?"<br> O length<br> volume
anygoal [31]
The answer is volume.
6 0
3 years ago
Which term is used to describe the reactant that is used up completely and controls the amount of product that can be produced d
givi [52]

Answer: Option (c) is the correct answer.

Explanation:

A limiting reagent is defined as a reagent that completely gets consumed in a chemical reaction. A limiting reagent limits the formation of products.

For example, we have given 5 mol of A and the reaction is 2A + 4B \rightarrow 2AB

Whereas when 4 mol B will react with 2 mol of A. Hence, 8 mol of B will react with 4 mol A as follows.

                        \frac{2}{4} \times 8 = 4 mol

As, the given moles of A is more than the required moles. Thus, it is considered as an excess reagent.

Hence, B is a limiting reagent because it limits the formation of products.

Thus, we can conclude that limiting reactant is the term used to describe the reactant that is used up completely and controls the amount of product that can be produced during a chemical reaction.  

6 0
3 years ago
HELP FAST FOR BRAINLIEST
krek1111 [17]

Answer:

only ones ik

14: dependant

18: constraints

3 0
3 years ago
The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.
adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol

Now we have to calculate the energy required.

Q=\frac{\Delta H}{n}

where,

Q = energy required

\Delta H = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

7 0
3 years ago
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