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Butoxors [25]
3 years ago
5

Please help i will mark brainlist

Chemistry
1 answer:
Mariulka [41]3 years ago
5 0

Answer:

432cm³

Explanation:

Given parameters:

Length  = 12cm

Height  = 6cm

Width  = 6cm

Unknown:

V = ?

Solution:

We have been given that;

     

                    V  = L x W x H

So;

                 V  = 12cm  x 6cm x 6cm  = 432cm³

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DEFINE THESE WORDS FOR A TON OF POINTS!!!!!!!
PilotLPTM [1.2K]
Hi , these are the definitions
Chemistry; the science that deals with the composition and properties of substances and various elementary forms of matter.
aquatic system ; ecosystem in a body of water , communities and organisms that are dependent on each other.
depth ; the distance from the top to the bottom of something .
geography ; the study of the earth s  physical features and the people ,plants , and animals.
light ; brightness from the sun or from a light.
medium ; between small and  large in   size.
period ;  the name given to a horizontal row of the periodic table.
properties ; attributes of a substance.
refraction ; the change of direction of a ray of light , sound , heat ....
reflection ; reflection is when infrared waves  bounce  from a surface .
salinity ; a saline substance or a liquid that contains salt.
pressure ; measure of the force applied over an  unit area .
photosynthesis ; the process used by plants , algae ,  and certain bacteria to harness energy from sunlight into chemical energy.
thermocline ; an abrupt  temperature gradient in a body of water .
intertidal ; relating to the region    between the hide tide mark and the low  tide mark .
benthic ; relating to the bottom of a sea or a lake or to the organisms that live there .
pelagic ; relating to or living in the sea far from the shore.
epipelagic ; relating to or inhabiting the uppermost layer of the water column of the open ocean  , into which enough sunlight enters for photosynthesis to take place.
mesopelagic ; relating to or inhabiting the layer of the water column in the open sea that lies between the epipelagic   and bathypelagic  layers at depths of about 200 to 1,000 meters.
bathypelagic ; relating to or inhabiting the layer of the water column  of the open sea that lies between the mesopelagic  and abyssopelagic  layers at depths of about 1,000 to 4,000.
abyssopelagic ; refering to or occurring in the region of deep water above the floor of the ocean .
neritic ; region of water lying  directly above the sub littoral zone of the sea bottom .
photic ;   pertaining to the generation of light by organisms .
aphotic; zone of an ocean , lowest level  at which photosynthesis can take place .
commensalism ; type of a relationship between a plant , an animal , fungus , etc.
competition ; the struggle among  organisms .
freshwater systems ; subset of earth s aquatic ecosystems .
marine systems ; inside aquatic system , marine systems include oceans , seas ,etc.
mutual ism ; relation between species of organisms  in which both benefit from the association.
organism ; a system with many parts that depends on each other and work together. 
parasitism ; the behavior of  a parasite .
predation; preying or plundering.
relationships ; the way in which two things are connected .
terrestrial systems ; an ecosystem only found on land forms.


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4 0
4 years ago
If 2 CH3OH + 3 O2 -&gt; 2CO2 + 4 H2O was carried out in the laboratory and 219 g of water was produced, what would the percent y
ANEK [815]

Answer:

Percent yield = 84.5 %

Explanation:

Given data:

Mass of methanol = 229 g

Actual yield of water = 219 g

Percent yield of water = ?

Solution:

Chemical equation:

2CH₃OH + 3O₂  →  2CO₂  + 4H₂O

Number of moles of methanol:

Number of moles = mass/ molar mass

Number of moles = 229 g/ 32 g/mol

Number of moles = 7.2 mol

Now we will compare the moles of water with methanol.

                        CH₃OH         :            H₂O

                            2               :               4

                           7.2             :           4/2×7.2 = 14.4 mol

Mass of water:

Mass = number of moles × molae mass

Mass = 14.4 mol × 18 g/mol

Mass = 259.2 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 219 g / 259.2 g × 100

Percent yield = 84.5 %

7 0
3 years ago
5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
3 years ago
Energy in a chemical reaction can be identified as which of the following?
daser333 [38]
I believe it’s A..but I’m not quite sure.
6 0
3 years ago
The following information was recorded by a student team working to prepare nickel sulfate. Plan: Prepare NiSO4 by reacting NiO
mixas84 [53]

Answer:

The options e and d are correct.

Explanation:

Mass of NiO = 7.5 g

Moles of NiO = \frac{7.5 g}{74.69 g/mol}=0.10 mol

Moles of sulfuric acid = n

Volume of sulfuric acid ,V= 50 mL = 0.050 L

Molarity of sulfuric acid ,M = 6 mol/L

n=M\time V=6mol/L\times 0.050 L =0.3 mol

NiO + H_2SO_4\rightarrow NiSO_4 + H_2O

According to reaction, 1 mole of NiO reacts with 1 mole of sulfuric acid.

Then 0.10 moles of NiO reacts with :

\frac{1}{1}\times 0.10 mol/=0.10 mol of sulfuric acid.

As we can see that sulfuric acid is in excess amount, so the amount of the product will depend upon amount of NiO.

According to reaction, 1 mole of NiO gives with 1 mole of NiSO_4.

Then 0.10 moles of NiO wil give :

\frac{1}{1}\times 0.10 mol/=0.10 mol of  NiSO_4.

Molar mass of  NiSO_4 = 154.75 g/mol

Mass of 0.10 moles of NiSO_4:

= 154.75 g/mol × 0.10 mol = 15.475 g

Theoretical mass of NiSO_4 = 15.475 g

Experimental yield of NiSO_4 = 17.4 g

Percentage yield :

\Yield=\frac{\text{Experimental mass}}{\text{Theoretical mass}}\times 100

Percentage yield of NiSO_4:

\Yield=\frac{17.4}{15.475 g}\times 100=112\%

Moles of NiSO_4.6H_2O = 262.85 g/mol × 0.10 mol = 26.285 g

Experimental yield of NiSO_4.6H_2O = 17.4 g

Percentage yield of NiSO_4.6H_2O:

\Yield=\frac{17.4}{26.285 g}\times 100=66.2\%

3 0
3 years ago
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