I'd say by Neutralization so as to make it Harmless.
Or
You Over dilute it with Water
Note: You pour the acid into water...Not water into acid as it may spill and burn your skin.
Answer: D
Explanation:
6. Heat from the sun reaches to us by
(a) radiation
(b) conduction
(c) convection
(d) all of these
PS. the one who gets it correct will be followed by me and also be marked as the brainliest!
Answer:
radiation
....
.
.
.
.hi how r u
Is my answer correct
Answer:
Radiation
Explanation:
Heat from the sun reaches is by the process of radiation
Which property altered during a chemical change is not altered during a physical change?
A. composition of the matter
B. temperature of the matter
C. volume of the matter
D. phase of the matter
The property altered during a chemical change that is not altered during a physical change is composition of the matter.
A physical change is one in which no new substance is formed and it is easily reversible.
A chemical change is one that is not easily reversible and no new substance is formed. It may be accompanied by absorption or evolution of heat.
The composition of a substance changes during a chemical change because bonds between atoms break and new bonds are formed. This does not occur during a physical change.
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What would most likely happen if the kangaroo rats
were killed off and removed from the food chain?
Answer:
Explanation:
The number of cactus plants would increase.
Am element has 18 electrons ,20 neutrons and a charge of -2.what is the mass number?..a.38 b.40 c.32 d.39
Answer:
Surely you must have a Periodic Table beside you....?
a=38
Explanation:
You gots Z=18, the atomic number, where Z is the number of nuclear protons. The number of protons defines the element....and thus we got calcium....
....and with 20 neutrons we got the 38Ca isotope....the which is rather short-lived...
Which pair of reactants for a Grignard reaction does not give 2-phenylbutan-2-ol after an aqueous workup?
3. If we remember seeing something happen, we can trust that it happened just as we think it did. True or False
[tex] \huge{ \underbrace{ \overbrace{True}}}[/tex]
.
[tex]\huge\mathfrak\purple{Hope \: it \: helps}[/tex]
Note : Because Many of My Dreams came true and I realized.
In nature, Chlorine-35 isotopes (35 amu) make up 75
% of all chlorine. Chlorine-37 (37 amu) isotopes make
up 25 % of all chlorine,
Show your calculation for the AVG. Atomic Mass of
chlorine:
What is a mixture? How will you separate the components of a mixture containing sugar, sand and water?
Answer:
Mixture is combination of two or more substances
Explanation:
The sugar would dissolve in water. You could then pour off the solution and wash the remaining sand with a bit more water. Heat the water to evaporate it from the sugar, and the two are separated.
How many inches are in 4.5 miles?
when two atoms share four valence electrons the shared electrons form a/an ____
Which of the following atoms is expected to have two unpaired electrons?
Al
Mg
Na
None of the above
Answer:
none of the above answers
calculate mass of MnO2 if 1 mole of KMnO4 completely burnt. 2KMnO4 -> K2MnO4+MnO2+O2
Answer: 80 grams
Explanation:
2KMnO4 -> K2MnO4+MnO2+O2
2 moles of KMnO4 would produce 1 mole of MnO2, so 1 mole of KMnO4 would produce 0.5 moles of MnO4. The molar mass of MnO4 is 160 g/mole. 1/2 mole of KMnO4 would therefore be (160 g/mole)*(0.5 moles) = 80 grams.
calculate the number of each atom in 2.5 gram of caco3
Answer:
[tex]2.5\; \rm g[/tex] of [tex]\rm CaCO_{3}[/tex] would contain:
Approximately [tex]1.5 \times 10^{22}[/tex] calcium atoms (approximately [tex]0.025\; \rm mol[/tex],)Approximately [tex]1.5 \times 10^{22}[/tex] carbon atoms (approximately [tex]0.025\; \rm mol[/tex],) andApproximately [tex]4.5 \times 10^{22}[/tex] oxygen atoms (approximately [tex]0.075\; \rm mol[/tex].)Explanation:
Look up the Avogadro constant: [tex]N_{\rm A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}[/tex].
For example, "[tex]1\; \rm mol[/tex] of carbon atoms" would contain [tex]N_{\rm A}[/tex] carbon atoms (approximately [tex]6.022\times 10^{23}[/tex]) by definition.
Look up the relative atomic mass of carbon, calcium, and oxygen on a modern periodic table:
Calcium: [tex]40.078[/tex].Carbon: [tex]12.011[/tex].Oxygen: [tex]15.999[/tex].In other words, the mass of [tex]1\; \rm mol[/tex] of calcium atoms would be [tex]40.078\; \rm g[/tex]. The mass of [tex]1\; \rm mol\![/tex] of carbon atoms would be [tex]12.011\; \rm g[/tex], and the mass of [tex]1\; \rm mol \!\![/tex] of oxygen atoms would be [tex]15.999\; \rm g[/tex].
As the formula [tex]\rm CaCO_{3}[/tex] suggests, every formula unit of this ionic compound includes one calcium atom, one carbon atom, and three oxygen atoms. The formula mass of [tex]\rm CaCO_{3}\![/tex] would give the mass of every mole of [tex]\rm CaCO_{3}\!\![/tex] formula units.
Calculate the formula mass of [tex]\rm CaCO_{3}[/tex] from the relative atomic mass data:
[tex]\begin{aligned} & M({\rm CaCO_{3}}) \\ =\; & 40.078\; \rm g \cdot mol^{-1} \\ & + 12.011\; \rm g \cdot mol^{-1} \\ & + 3 \times (15.999\; \rm g \cdot mol^{-1}) \\ =\; & 100.086\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Calculate the number of [tex]\rm CaCO_{3}[/tex] formula units in that [tex]2.5\; \rm g[/tex] of this compound:
[tex]\begin{aligned}& n({\rm CaCO_{3}}) \\ =\; & \frac{m({\rm CaCO_{3}})}{M({\rm CaCO_{3}})} \\ =\; & \frac{2.5\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \\ \approx\; & 0.025\; \rm mol\end{aligned}[/tex].
In other words, [tex]2.5\; \rm g[/tex] of [tex]\rm CaCO_{3}[/tex] would contain approximately [tex]0.025\; \rm mol[/tex] [tex]\rm CaCO_{3}\![/tex] formula units.
Again, there are one calcium atom, one carbon atom, and one oxygen atom in every [tex]\rm CaCO_{3}[/tex] formula unit. That approximately [tex]0.025\; \rm mol[/tex] [tex]\rm CaCO_{3}\![/tex] formula units would thus contain:
Approximately [tex]1 \times 0.025\; \rm mol = 0.025\; \rm mol[/tex] calcium atoms, Approximately [tex]1 \times 0.025\; \rm mol = 0.025\; \rm mol[/tex] carbon atoms, andApproximately [tex]3 \times 0.025\; \rm mol = 0.075\; \rm mol[/tex] oxygen atoms.Make use of the Avogadro constant to convert the numbers.
For example, the number of calcium atoms in that approximately [tex]0.025\; \rm mol[/tex] of calcium atoms would be:
[tex]\begin{aligned} & N({\text{calcium}) \\ = \; & n({\text{calcium}) \cdot N_{\rm A} \\ \approx \; & 0.025\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 1.5 \times 10^{22} \end{aligned}[/tex].
Likewise, the number of carbon atoms in that approximately [tex]0.025\; \rm mol[/tex] of carbon atoms would be:
[tex]\begin{aligned} & N({\text{carbon}) \\ = \; & n({\text{carbon}) \cdot N_{\rm A} \\ \approx \; & 0.025\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 1.5 \times 10^{22} \end{aligned}[/tex].
The number of oxygen atoms in that approximately [tex]0.075\; \rm mol[/tex] of oxygen atoms would be:
[tex]\begin{aligned} & N({\text{oxygen}) \\ = \; & n({\text{oxygen}) \cdot N_{\rm A} \\ \approx \; & 0.075\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 4.5 \times 10^{22} \end{aligned}[/tex].
Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
bubble and foam!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
3(b) The balloon did not burst when it is pressed onto 100 nails because
1. the nails are not sharp
2.the surface area of 100 nails produce high pressure on the balloon
3. the larger surface area of 100 nails produce low pressure on the balloon
3(c) To reduce pressure on the contact surface, we must apply force onto the .................... surface area.
1.largest
2.smallest
3. Any size of
b. Use the balanced equation to answer the following questions.
CuSO4(aq) + 2NaOH(aq) —->Cu(OH)2(s) + Na2SO4(aq)
i. What is the ratio of moles of CuSO4 to moles of NaOH?
ii. If 638.44 g CuSO4 reacts with 240.0 NaOH, which is the limiting reagent?
iii. Using the limiting reagent to determine how many grams of Cu(OH)2 should precipitate out in the reaction.
iv. If only 174.6g of Cu(OH)2 precipitate were actually collected from the reaction, what would the percent yield be?
Answer:
i. 1 : 2
ii. NaOH is the limiting reagent
iii. 292.5g
iv. 59.69%
Explanation:
I have the detailed and self-explanatory workings. will snap and post later. Battery percent at 15%, flashlight not working.
1. The ratio of CuSO₄ to NaOH is 1 : 2
2. The limiting reactant is NaOH.
3. The mass of Cu(OH)₂ that will precipitate out is 292.5 g.
4. The percentage yield of Cu(OH)₂ is 59.7%
1. Determination of the ratio of CuSO₄ to NaOH.
CuSO₄ + 2NaOH —> Cu(OH)₂ + Na₂SO₄
From the balanced equation above, we can see that the ratio of CuSO₄ to NaOH is 1 : 2
2. Determination of the limiting reactant.
CuSO₄ + 2NaOH —> Cu(OH)₂ + Na₂SO₄
Molar mass of CuSO₄ = 63.5 + 32 + (16×4) = 159.5 g/mol
Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Mass of NaOH from the balanced equation = 2 × 40 = 80 g
From the balanced equation above,
159.5 g of CuSO₄ reacted with 80 g of NaOH.
Therefore,
638.44 g of CuSO₄ will react with = (638.44 × 80) / 159.5 = 320.22 g of NaOH.
From the calculations made above, we can see that a higher mass (i.e 320.22 g) of NaOH than what was given (i.e 240 g) is needed to react completely with 638.44 g of CuSO₄.
Therefore, NaOH is the limiting reactant.
3. Determination of the mass of Cu(OH)₂ that will precipitate out.
CuSO₄ + 2NaOH —> Cu(OH)₂ + Na₂SO₄
Molar mass of Cu(OH)₂ = 63.5 + 2(16 + 1) = 97.5 g/mol
Mass of Cu(OH)₂ from the balanced equation = 1 × 97.5 = 97.5 g
From the balanced equation above,
80 g of NaOH reacted to produce 97.5 g of Cu(OH)₂.
Therefore,
240 g of NaOH will react to produce = (240 × 97.5) / 80 = 292.5 g of Cu(OH)₂
Thus, 292.5 g of Cu(OH)₂ precipitated out of the reaction.
4. Determination of the percentage yield.
Actual yield of Cu(OH)₂ = 174.6 gTheoretical yield of Cu(OH)₂ = 292.5 gPercentage yield of Cu(OH)₂ =?Percentage yield = (Actual /Theoretical) × 100
Percentage yield of Cu(OH)₂ = (174.6 / 292.5) × 100
Percentage yield of Cu(OH)₂ = 59.7%
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What kinds of questions cannot be answered by chemistry
In a year an adult cow produces around 3.571 x 10^27 molecules of methane CH4 a greenhouse gas how many moles of methane does this cow produce in a year
Answer:
[tex]\boxed{\boxed {\sf 5930.0 \ mol \ CH_4}}[/tex]
Explanation:
We are asked to find how many moles of methane (CH₄) a cow produces in a year. We must convert molecules of methane to moles of methane. We will do this using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance.
In this case, the particles are molecules of methane. There are 6.022 × 10²³ molecules of methane in 1 mole.
We will convert molecules to moles using dimensional analysis. Set up a ratio using Avogadro's Number.
[tex]\frac {6.022 \times 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]
We are converting 3.571 × 10²⁷ molecules of methane to moles, so we multiply the ratio by this value.
[tex]3.571 \times 10^ {27} \ molecules \ CH_4 *\frac {6.022 \times 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]
Flip the ratio. It is still the same value, but the units of molecules of methane cancel.
[tex]3.571 \times 10^ {27} \ molecules \ CH_4 *\frac { 1 \ mol \ CH_4}{6.022 \times 10^{23} \ molecules \ CH_4}[/tex]
[tex]3.571 \times 10^ {27} *\frac { 1 \ mol \ CH_4}{6.022 \times 10^{23} }[/tex]
[tex]\frac {3.571 \times 10^ {27}}{6.022 \times 10^{23} } \ mol \ CH_4[/tex]
[tex]5929.923613 \ mol \ CH_4[/tex]
The original measurement of molecules of methane( 3.571 × 10²⁷ molecules) has 4 significant figures, so our answer must have the same. For the number we calculated, that is the ones place. The 9 in the tenth place tells us to round the 9 in the ones place up to a 0, then the 2 in the tens place up to a 3.
[tex]5930 \ mol \ CH_4[/tex]
We must add a placeholder zero in the tenth place to reach 4 sig figs.
[tex]5930.0 \ mol \ CH_4[/tex]
The adult cow produces approximately 5930.0 moles of methane in a year.
for the equation given, how many grams of methane will react with a 125g of oxygen
CH4 (g) + 2O2(g) -> CO2(g) +2H2O
Answer:
16
Explanation:
1) in the given equation CH₄+2O₂⇒ 2H₂O+CO₂;
M(CH₄)=12+4=16 (g/mol); M(O₂)=32 (g/mol);
2) if m(O₂)=125 gr., then ν(O₂)=m(O₂)/M(O₂)=125/32≈3.90625 (moles);
3) if in the given equation 'ν' of O₂ is 2, then the 'ν' of CH₄ is 1;
4) m(CH₄)=ν(CH₄)*M(CH₄)=1*16=16 (gr.).
why does copper Cu (I) have a short lifespan
Answer:
Copper (Cu), chemical element, a reddish, extremely ductile metal of Group 11 (Ib) of the periodic table that is an unusually good conductor of electricity and heat. Copper is found in the free metallic state in nature.
Explanation:
hope this helps:)
11) Calcium Chloride, CaCl2, can be used instead of road salt to melt the ice on roads during the winter. To determine how much calcium chloride had been used on a nearby road, a student took a sample of slush to analyze. The sample had a mass of 23.47g. When the solution was evaporated, the residue had a mass of 4.58g (assume that no other solutes were present).
a) What was the mass/mass percent of calcium chloride in the slush
b) How many grams of calcium chloride were present in 100g solution?
Answer:
What is that? I do not understand
How can the Swartz process be modified to work for bacteria? Explain.
Answer:
https://swartzbiotechnologylab.sites.stanford.edu/research
Explanation:
Alkyl chlorides and alkyl bromides are typically transformed into alkyl fluorides in the organic reaction known as the Swarts Reaction.
Swartz reaction?Frederick Jean Edmond Swarts first reported this procedure in 1892. Heating the alkyl chloride or alkyl bromide in the presence of heavy metal fluorides like AGF, Hg2F2, CoF2, or SbF3 causes this reaction.
Swartz process can be used to modify work for bacteria. Bacterial biofilms are collections of microorganisms where the cells are immersed in an extracellular polymeric matrix that the bacteria have self-produced.
In contrast to planktonic bacteria, which are in a state of free movement in a bulk solution, this situation is very different.
Applications:
In this process, alkyl fluorides are created.Fluorinated aliphatic organic molecules, or Freons, are produced via a different variation of the same process.Anhydrous hydrogen fluoride is fluorinated to produce Freons in the presence of antimony salts, which exhibit the oxidation states +3 and +5.Therefore, Alkyl chlorides and alkyl bromides are typically transformed into alkyl fluorides in the organic reaction known as the Swarts Reaction.
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the internacional system of units is based on ?
Answer:
The international system of units is basic on 7 units.
Hope it helps you.Which of the following is a feature of a Type IV flotation device?
Answer:
A Type IV PFD is an approved device designed to be thrown to a person in the water. It is not designed to be worn. It is designed to have at least 16.5 pounds of buoyancy. The most com- mon Type IV PFD is a buoyant cushion.
Hello please I need help
Illustrate the law of law of multiple proportions using the formation of SO2 and SO3 by sulphur and oxygen
As illustrated from law of multiple proportion, the mass of the oxygen which combined with a fixed mass of the sulphur is in simple whole number ratio of 2 is to 3 (2:3).
The law of Multiple proportion states that when two elements combine to form more than one compound, the mass of the second element, which combines with a fixed mass of the first element, will always be ratios of small whole numbers.
To illustrate the law of multiple proportion from sulphur and oxygen, we use the given compounds.
The reaction of oxygen with sulphur to form SO₂ and SO₃:
[tex]S \ + \ O_2 \ -----> SO_2\\\\2SO_2 \ + \ O_2 \ ---> \ 2SO_3[/tex]
atomic mass of oxygen = 16 g
atomic mass of sulphur = 32 g
In the formation of SO₂:mass of sulphur = 32 g and mass of oxygen = (2x16) = 32 g
In the formation of SO₃:mass of sulphur = 32 g and mass of oxygen = (3 x 16) = 48 g
Notice, while mass of sulphur is constant, mass of oxygen changed.
The ratio of mass of oxygen in the given two compounds = 32g : 48g = 2:3
Thus, based on the law of multiple proportion, the mass of the oxygen which combined with a fixed mass of the sulphur is in simple whole number ratio of 2 is to 3 (2:3).
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what is seed reproduction
Answer:
A seed is formed when fertilised ovule divides by mitosis. ... The ovary of the flower develops into the fruit while ovules develop into seeds. The formation of seed completes the process of reproduction in plants. Within the seed, the growing embryo develops and matures.
I hopes it helps you.Neha and Reha are playing see-saw. Neha is sitting 60 cm away from the fulcrum and Reha is sitting 40 cm away from the fulcrum. Calculate the effort that Reha should apply to lift Neha. The weight of Neha is 360 N.
Answer:
Reha should apply 540 N to lift Neha.
Explanation:
L = 360 N
LD = 60 m
E =?
ED = 40m
NOW
L * LD = E * ED
360 *60 = E*40
21600/40 = E
540 N = E
A graduated cylinder is filled with 20ml of water. A rock is dropped in to the graduated cyclinder and the volume of water rises to 30ml
inside the graduated cylinder. The mass of the rock is 23g. What is the density of the rock?
2.3g/ml
0.77g/ml
5g/ml
1.15g/ml
Answer:
[tex]\boxed {\boxed {\sf 2.3 \ g/mL}}[/tex]
Explanation:
We are asked to find the density of a rock. Density is the substance's mass per unit volume. The formula for calculating density is:
[tex]\rho= \frac{m}{v}[/tex]
The mass of the rock is 23 grams. The volume was found using water displacement. A known amount of water was measured, the rock was added to the graduated cylinder, and the water level was recorded again. The volume is the difference between the final and initial water level. The graduated cylinder originally had 20 milliliters and the water rose to 30 milliliters.
volume = final water level - initial water level volume = 30 mL - 20 mLvolume = 10 mLNow we know the mass and the volume.
m= 23 g v= 10 mLSubstitute the values into the formula.
[tex]\rho= \frac{ 23 \ g }{30 \ mL}[/tex]
Divide.
[tex]\rho= 2.3 \ g/mL[/tex]
The density of the rock is 2.3 grams per milliliter.
identify 2 ways to measure mass
Answer:
The two ways to measure mass are subtraction and taring.
