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3 years ago

What is the scientific tool required to measure the volume of a liquid?

Chemistry

1 answer:

Vedmedyk [2.9K]3 years ago

3 0

Volumetric flask? Graduated cylinder? micro-pipets? There's a lot of tools that measure volume

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Oxygen is a __________ and nitrogen is a __________. metalloid, metalloid nonmetal, metal nonmetal, nonmetal nonmetal, metalloid

Anna35 [415]

Answer:

"nonmetal, nonmetal"

Explanation:

Oxygen is a non metal and Nitrogen is a non metal. It is 8th element of the periodic table. It is located in period 2 and group 16.

Nitrogen lies at the group 15 of the periodic table. Its atomic no is 7. Its valency is 2.

Hence, the correct option is (c) "nonmetal, nonmetal".

4 0

3 years ago

How many grams of O2(g) are needed to completely burn 15.6 g of C3H8(g)?

alexdok [17]

15.6gC3H8*44.1gC3H8*32gO2=11.32

7 0

4 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago

Which wave has the higher frequency? <br><br> A. Bottom<br> B. Top

ladessa [460]

The top one does because there are more and it’s closer

7 0

3 years ago

When a Na atom loses an electron to a Cl atom, the two atoms have opposite charges and are attracted to one another. What is the

ahrayia [7]

Answer:

Ionic bond

Explanation:

When Na loses an electron, it becomes positive ion.

When Cl gains an electron, it becomes negative ion.

The electrostatic attraction between these ions is an ionic bond.

7 0

3 years ago