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Alona [7]
3 years ago
7

Which of the following are observations? The candle floats. The tires are flat. The dog is barking. The tire is flat because of

a decrease in temperature. If a person walks by, then the dog will bark.
Chemistry
2 answers:
matrenka [14]3 years ago
8 0

Answer:

The first three statements, that is, the candle floats, the tires are flat, and the dog is barking are observations.

Explanation:

Observation refers to the process or action of observing someone or something cautiously in order to attain a certain kind of information. Thus, the first three given statements are examples of observations, while the last two statements imply the concept of cause-effect relationships. The cause and effect refer to an association between the things or the incidents, where one is the outcome of the other or the others. This is an amalgamation of action and reaction.

Liula [17]3 years ago
6 0
Everything but "<span>The tire is flat because of a decrease in temperature" is an observation.</span>
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Consider the reaction: CH4 + 2O2 = CO2 + 2H2O
crimeas [40]

Answer:

The answer to your question is:

a)  80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

                               x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

                               x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

                                 CH4   +   2O2   ⇒   CO2   +   2H2O

                        64 g of O2 ------------------  44 g of CO2

                        22 g of O2 ------------------   x

                        x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________  

                           CH4   +   2O2   ⇒   CO2   +   2H2O                              

                           10.31 g                     20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0
3 years ago
Which one of the following formulas represents an aldehyde? 
Masteriza [31]
The formula of Aldehyde is represented by <span>C2H4O. It has two atoms of Carbon, four atoms of hydrogen and one atom of oxygen. Aldehyde is an organic compound. It's organic because it contains carbon. It has a structure of R-CHO, that consists a carbonyl center bonded to R group and to hydrogen.</span>
8 0
3 years ago
What is the momentum of a 1,700 kg car travelling in a straight line at 13 m/s?
Anna71 [15]

Answer:

Explanation:

We know that momentum is the product of mass and velocity so here

mass (m) = 1700 kg

velocity (v) = 13 m/s

So now

momentum = m * v

                   = 1700 * 13

                   = 22100 kg m/s

hope it helps :)

4 0
3 years ago
Ethylene produced by fermentation has a specific gravity of 0.787 at 25 degree Celsius. What is the volume of 125g of ethanol at
WITCHER [35]

<u>Answer:</u> The volume of given amount of ethanol at this temperature is 159.44 mL

<u>Explanation:</u>

Specific gravity is given by the formula:

\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}

We are given:

Density of water = 0.997 g/mL

Specific gravity of ethanol = 0.787

Putting values in above equation, we get:

0.787=\frac{\text{Density of a substance}}{0.997g/mL}\\\\\text{Density of a substance}=(0.787\times 0.997g/mL)=0.784g/mL

Density is defined as the ratio of mass and volume of a substance.

\text{Density}=\frac{\text{Mass}}{\text{Volume}} ......(1)

Given values:

Mass of ethanol = 125 g

Density of ethanol = 0.784 g/mL

Putting values in equation 1, we get:

\text{Volume of ethanol}=\frac{125g}{0.784g/mL}=159.44mL

Hence, the volume of given amount of ethanol at this temperature is 159.44 mL

6 0
3 years ago
Which of the following are true statements about equilibrium systems? For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(
Grace [21]

Answer:

The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

⇒ Le Chatellier says As the CaCO3 concentration is increased, the system will attempt to undo that concentration change by shifting the balance to the right. <u>This statement is true.</u>

<u />

2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

⇒ Le chatellier says that if we increase the pressure, the equilibrium will shift to the side with the least number of particles.

Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. <u>This statement is false. </u>

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

⇒ Le Chatellier says if we increase the pressure, the equilibrium will shift to the side with the most particles.

In this case we have 2 moles of H2 and 1 mole of O2 on the left side and 2 mole of H2O on the right side. This means on the left side are more particles. So the equilibrium will shift to the left, so <u>this statement is true.</u>

4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

⇒ Le chatellier says if H2 will be removed (this means the left side will get less particles) so the equilibrium will shift to the left, to increase the amount of F2.

⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

If we increase the temperature of an endotherm reaction, the equilibrium will shift to the side of the products (the right side). <u>This statement is true.</u>

4 0
3 years ago
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