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3 years ago

Which statement describes how chemical formula such as H20 represent compounds check all that apply

Chemistry

1 answer:

Allisa [31]3 years ago

8 0

<h3>Answer:</h3>

They show the elements of make up a compound
They show the types of atoms that make up a molecule
They show the number of each type of atom in a molecule

<h3>Explanation:</h3>

To answer this question we need to know what is a compound.

A compound is a substance that is made by two or more atoms of different elements that are chemically combined.
A chemical formula of a compound is a formula that shows the elements that make up a compound and also the types of atoms that make up the molecule.
Additionally, the chemical formula shows the number of each type of atom in a molecule.
In this case, the chemical formula of a molecule of H₂O shows that it is made up of one hydrogen atom and two oxygen atoms.

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What is required in order to determine wether or not an object moves

Stels [109]

Answer:

The answer would be Point of Reference.

In order to determine, whether an object has moved or not, a point of reference is required. The reference point is required to the measure the movement of the object, without a reference point the distance or displacement cannot be obtained. Hence, a reference point is very required to determine, whether the object has moved or not.

(Hope this helps) Sky

7 0

3 years ago

How is oxide different from a neutral oxygen atom

Zepler [3.9K]

Answer: The main difference between oxide and oxygen is that oxide is a chemical compound with at least one oxygen atom while oxygen is an element whose atomic number is 8.

Explanation: let me know if it was right or wrong

4 0

3 years ago

A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe

Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: $V_{1}$ = 571 mL, $T_{1} = 26^{o}C$

(a) $T_{2} = 5^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL$

(b) $T_{2} = 95^{o}F$

Convert degree Fahrenheit into degree Cesius as follows.

$(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL$

(c) $T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL$

8 0

3 years ago

What is the change in density if a sample goes from 3.21 g/L to 5.43 g/mL?

Step2247 [10]

Answer:

$\Delta \rho =2.22 g/mL$

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

$\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL$

Best regards.

3 0

3 years ago

A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu

bogdanovich [222]

Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

Explanation:

The temperature of an object can be written using different temperature scales.

The two most important scales are:

- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).

- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

$T(K)=T(^{\circ}C)+273.15$

Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

6 0

4 years ago