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Answer:

320 g.

Explanation:

Hello!

In this case, according to the balanced chemical reaction, we can compute the grams of copper I chloride produced by each reactant, as shown below:

[tex]m_{CuCl}^{by\ Cu}=3.23molCu*\frac{2molCuCl}{2molCu}*\frac{99.0gCuCl}{1molCuCl} =320gCuCl\\\\m_{CuCl}^{by\ Cl_2}=1.64molCl_2*\frac{2molCuCl}{1molCl_2}*\frac{99.0gCuCl}{1molCuCl} =325gCuCl[/tex]

Thus, since copper produces the fewest grams of CuCl, we infer it is the limiting reactant, therefore the correct mass of copper I chloride is 320 g.

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B

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Answer:

EARTH

Explanation:

Earth‘s gravity, as already noted, is equivalent to 9.80665 m/s² (or 32.174 ft/s²). This means that an object, if held above the ground and let go, will accelerate towards the surface at a speed of about 9.8 meters for every second of free fall.

Answer:

[tex]Y=34.3\%[/tex]

Explanation:

Hello there!

In this case, according to the balanced chemical reaction:

[tex]2Al + Cr_2O_3 \rightarrow Al_2O_3 + 2Cr[/tex]

We notice there is a 1:1 mole ratio between Al2O3 and Cr2O3; thus, the following stoichiometric setup is used to compute the theoretical yield first:

[tex]m_{Al_2O_3 }=23.3gCr_2O_3*\frac{1molCr_2O_3}{151.99gCr_2O_3} *\frac{1molAl_2O_3}{1molCr_2O_3} *\frac{101.96gAl_2O_3}{1molAl_2O_3} \\\\m_{Al_2O_3}=15.6gAl_2O_3[/tex]

Thus, the percent yield turns out:

[tex]Y=\frac{5.35g}{15.6g} *100\%\\\\Y=34.3\%[/tex]

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Answer:

[tex]V=5.2 mL=0.052L[/tex]

Explanation:

Hello!

In this case, since the chemical reaction between silver nitrate and sodium chloride is:

[tex]AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)[/tex]

We can see there is a 1:1 mole ratio between each solution; thus, we first compute the moles of each reactant considering their molar masses:

[tex]n_{AgNO_3}=14.77g*\frac{1mol}{169.87g}=0.087molAgNO_3\\\\ n_{NaCl}=0.2631g*\frac{1mol}{58.44}=0.0045molNaCl[/tex]

Now, since the concentration of the silver chloride solution is 0.087 M, we may assume that the concentration of the NaCl solution is the same, so we can compute the volume as shown below:

[tex]V=\frac{n_{NaCl}}{M}=\frac{0.0045mol}{0.087mol/L}\\\\V=0.052L[/tex]

Or:

[tex]V=5.2 mL[/tex]

Best regards!

The volume of solution needed to react with 0.2631 g of NaCl  is 0.052 L.

Volume of the solution will be calculated by using the below formula:

M = n/V, where

M = concentration in terms of molarity

n = no. of moles

V = volume

Given chemical reaction is:

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

First we calculate the moles of given reactants by using the formula:

n = W/M , where

W = given mass

M = molar mass

Moles of AgNO₃ = 14.77g / 169.87g/mole = 0.087 mole

Moles of NaCl = 0.2631g / 58.44g/mole = 0.0045 mole

Concentration of AgNO₃ = 0.087 mole / 1L = 0.087M

From the stoichiometry of the reaction it is clear that mole ration of AgNO₃ & NaCl is 1:1. So, we take the concentration of NaCl is equal to the concentration of AgNO₃ and calculate the volume by using the above formula as:

Volume of NaCl = 0.0045mole / 0.087M = 0.052 L

Hence, 0.052 L is the required volume of NaCl.

To know more about moles, visit the below link:

https://brainly.com/question/17199947

Answer:

Most manure consists of animal feces; other sources include compost and green manure. Manures contribute to the fertility of soil by adding organic matter and nutrients, such as nitrogen, that are utilised by bacteria, fungi and other organisms in the soil.

Explanation:

Hope it is helpful....

Answer:

Animal manure, such as chicken manure and cow dung, has been used for centuries as a fertilizer for farming. It can improve the soil structure (aggregation) so that the soil holds more nutrients and water, and therefore becomes more fertile. ... Manure is also commercially composted and bagged and sold as a soil amendment.

Explanation:

Answer:

22 g

Explanation:

Step 1: Write the balanced equation

N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)

Step 2: Calculate the moles of NH₃ produced from 2.0 moles of H₂

The molar ratio of H₂ to NH₃ is 3:2.

2.0 mol H₂ × 2 mol NH₃/3 mol H₂ = 1.3 mol NH₃

Step 3: Calculate the mass corresponding to 1.3 moles of NH₃

The molar mass of NH₃ is 17.03 g/mol.

1.3 mol × 17.03 g/mol = 22 g

Answer:

O + O ---Δ----> O2

Explanation:

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Answer:

hello my ne is yash follow me please

Answer: kinetic energy

Answer:

Kinetic Energy.

Explanation:

All these "forces" rely on specific motion.

Answer:

C)

Explanation:

Answer:

47.8 moles of H₂O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2H₂ + O₂ —> 2H₂O

From the balanced equation above,

1 mole of O₂ reacted to produce 2 moles of H₂O

Finally, we shall determine the number of mole of water, H₂O, produced by the reaction of 23.9 moles of O₂. This can be obtained as follow:

From the balanced equation above,

1 mole of O₂ reacted to produce 2 moles of H₂O.

Therefore, 23.9 moles of O₂ will react to produce = 23.9 × 2 = 47.8 moles of H₂O.

Thus, 47.8 moles of H₂O were obtained from the reaction.

Answer:

[tex]m_{Hg}=829.4gHg[/tex]

Explanation:

Hello there!

In this case, considering the Avogadro's number, which helps us to realize that 1 mole of mercury atoms contains 6.022x10²³ atoms and at the same time 1 mole of mercury weights 200.59 g, we obtain:

[tex]m_{Hg}=2.49x10^{24}atoms*\frac{1mol}{6.022x10^{23}atoms} *\frac{200.59g}{1molHg}\\\\m_{Hg}=829.4gHg[/tex]

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Answer:

If your asking what golds natural state of matter is it's solid.

Explanation:

Answer:

the answer is soild

Explanation:

i did it on edge :)

Answer:

Which infection may have fixes created because of their comprehension of construction and capacity of protein?

Answer:

NH₃

M = n/V(L)

0.844 mol (Both numbers have 3 significant figures so the result has 3 significant figures as well)

Explanation:

Step 1: Given and required data

Step 2: Calculate the moles (n) of ammonia (solute)

Molarity is equal to the moles of solute divided by the liters of solution.

M = n/V(L)

n = M × V(L)

n = 2.25 mol/L × 0.375 L = 0.844 mol (Both numbers have 3 significant figures so the result has 3 significant figures as well)

Answer:

relation (but not a function)

Explanation:

c on edge

Answer: I believe the answer is D.) All of the above.

Explanation:

Answer:

64.7

Explanation:

Answer:

64.7 is the answer

hope this helps

have a good day :)

Explanation:

answer:

a mountain ridge which circles the earth

subduction zones

Answer:

1.99×10¯¹⁴ M

Explanation:

We'll begin by writing the balanced dissociation equation of pyridine. This is illustrated below:

C₅H₅N + H₂O <=> C₅H₆N⁺ + OH¯

From the balanced equation above,

1 mole of C₅H₅N produced 1 mole of OH¯.

Therefore, 0.502 M C₅H₅N will also produce 0.502 M OH¯.

Finally, we shall determine the concentration of hydronium ion, H₃O⁺ in the solution. This can be obtained as follow:

Concentration of Hydroxide ion [OH¯] = 0.502 M

Concentration of hydronium ion [H₃O⁺] =?

[H₃O⁺] [OH¯] = 1×10¯¹⁴

[H₃O⁺] × 0.502 = 1×10¯¹⁴

Divide both side by 0.502

[H₃O⁺] = 1×10¯¹⁴ / 0.502

[H₃O⁺] = 1.99×10¯¹⁴ M

Thus, the concentration of hydronium ion, H₃O⁺ in the solution 1.99×10¯¹⁴ M

Answer:

B. physical property is the answer

Answer:

BC = 10.28 ≈ 10.3 mi to the nearest tenth

Explanation:

Using the trigonometry rules; SOH CAH TOA,

TOA would be more suitable in this question.

Tan ∅ = Opposite / Adjacent

Tan 61 = BC / AB

1.8040 = BC / 5.7

BC =  5.7 * 1.8040

BC = 10.28 ≈ 10.3 mi

the answer is 10.3, i already rounded it.

Answer:

[tex]V=3.50x10^{-22}L[/tex]

Explanation:

Hello there!

In this case, given the dimensions of the box, we first compute the volume by multiplying each side:

[tex]V=4nm*8.75nm*10nm=350nm^3[/tex]

Next, we apply the following conversion factor in order to obtain the corresponding liters:

[tex]V=350nm^3*(\frac{1m}{10^9nm} )^3*\frac{1000L}{1m^3} \\\\V=350nm^3*\frac{1m^3}{10^{27}nm^3} *\frac{1000L}{1m^3}\\\\V=3.50x10^{-22}L[/tex]

Best regards!

Answer:

Germanium monoxide

hope this helps :)

Answer:

2.0 × 10⁴ mg

Explanation:

Step 1: Given data

Mass (m): 0.020 kg

Step 2: Convert the mass from kilograms to grams

We will use the conversion factor 1 kg = 1,000 g.

0.020 kg × (1,000 g/1 kg) = 20 g

0.020 kg is equal to 20 g.

Step 3: Convert the mass from grams to milligrams

We will use the conversion factor 1 g = 1,000 mg.

20 g × (1,000 mg/1 g) = 2.0 × 10⁴ mg

20 g is equal to 2.0 × 10⁴ mg. Then, 0.020 kg is equal to 2.0 × 10⁴ mg.

Answer:

See explanation below

Explanation:

In this case, we are having a reaction between an anion and alkyl halide. The carbon 1 of the anion will act as nucleophile and will attack the electrophile, which is 5 carbon chain with the bromine in the third carbon.

Now, the nucleophyle is an alkyne of 3 carbon. According to the description, it should be:

CH₃ - C ≡ C⁻

And the alkyl halide is:

CH₃ - CH₂ CH(Br) - CH₂ - CH₃

And the final product after the reaction would be the following:

CH₃ - C ≡ C - CH - (CH₂CH₃)₂

However, in the attached picture you can see this better and the mechanism of reaction.

Hope this helps

Answer: Thus 24.0 g of [tex]SO_2[/tex] would be needed.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]      

[tex]\text{Moles of} O_2=\frac{6.00g}{32g/mol}=0.1875moles[/tex]

[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(l)[/tex]  

According to stoichiometry :

1 mole of [tex]O_2[/tex] require = 2 moles of [tex]SO_2[/tex]

Thus 0.1875 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.1875=0.375moles[/tex]  of [tex]SO_2[/tex]  

Mass of [tex]SO_2=moles\times {\text {Molar mass}}=0.375moles\times 64g/mol=24.0g[/tex]

Thus 24.0 g of [tex]SO_2[/tex] would be needed to completely react with 6.00 g of [tex]O_2[/tex] such that all reactants could be consumed.

Answer:

biosphere and lithosphere

Explanation:

The biosphere is described as the zone of life on Earth. It is a sum of all ecosystems. The biosphere is composed of living organisms and non-living factors.

The lithosphere is the outer part of the Earth such that this part is rocky. The lithosphere is made up of the brittle crust.

In general, chemicals enter Ecosystems through the biosphere and lithosphere.

A (they are experimentally determined exponents)