4 NH3 +5 02 +4NO + 6H2O
When 70.25 grams of NH3 react, how many moles of H20 are produced?

Answer:

[tex]m_{H_2O}=74.8gH_2O[/tex]

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible to realize there is a 1:2 mole ratio of sulfuric acid to water; thus, given the mass of the former and its molar mass (98.07 g/mol), it is possible to determine the mass of produced water as shown below:

[tex]m_{H_2O}=203.50gH_2SO_4*\frac{1molH_2SO_4}{98.07gH_2SO_4}*\frac{2molH_2O}{1molH_2SO_4} *\frac{18.02gH_2O}{1molH_2O}\\\\m_{H_2O}=74.8gH_2O[/tex]

Regards!

Answer:

ANSWER is E

Explanation:

2H₃PO₄ + 3Ca(OH)₂→ 6H₂O + Ca₃(PO₄)₂

Answer:

First step is to find what the products would be. Since this seems to be a double displacement, you get: H3PO4(aq)+Ca(OH)2(aq)→ H2O(L)+Ca3(PO4)2(s)

You know the products because H has a charge of +1, so PO4 must have a charge of -3, and since OH has a charge of -1, Ca must be +2. Then make each compound have a net charge of 0. H20 is a liquid since aq solutions have liquid water. Ca3(PO4)2 is a solid, use the solubility rules.

Lastly, balance the equation: 2H3PO4(aq)+3Ca(OH)2(aq)→ 6H2O(L)+Ca3(PO4)2(s)

So the salt formed is Ca3(PO4)2

Answer: 75.7 % yield

Explanation:

The balanced chemical equation is:

[tex]2H_2S+SO_2\rightarrow 3S+2H_2O[/tex]  

According to stoichiometry :

68.2 g of [tex]H_2S[/tex] will require = 64 g of [tex]SO_2[/tex]

Thus 7.5 g of [tex]H_2S[/tex] will require = [tex]\frac{64}{68.2}\times 7.5=7.0g[/tex] of [tex]SO_2[/tex]

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

As 68.2 g of give = 96 g of

Thus 7.5 g of [tex]H_2S[/tex] give =[tex]\frac{96}{68.2}\times 7.5=10.5g[/tex]  of [tex]S[/tex]

% yield=[tex]\frac{\text{Actual yield}}{\text {Theoretical yield}}\times 100=\frac{7.95g}{10.5g}\times 100=75.7\%[/tex]

Thus 75.7 % yield is there.

Answer:

laboratory use the organic substances dissolve in water methanol or ethanol then Sinha to rise iron chloride solution is added a transient permanent coloration usually purple green or blue indicate the presence of a funnel or an hall

Balanced equation for Fe reacting with [tex]Cl_2[/tex]

[tex]2Fe + 3Cl_2[/tex] → [tex]2FeCl_3[/tex]

Iron filings are small shavings of ferromagnetic material.

When hydrochloric acid (HCl) reacts with iron filings (Fe) a dissolving metal reaction occurs with the Fe being converted to iron chloride ([tex]FeCl_2[/tex]) and hydrogen ([tex]H_2[/tex]).

The balanced equation is:

[tex]Fe + 2HCl[/tex] → [tex]FeCl_2 + H_2[/tex]

Note that reaction only produces [tex]FeCl_2[/tex] and not [tex]FeCl_3[/tex]. To make [tex]FeCl_3[/tex] you have to get more forcing and react Fe with chlorine gas ([tex]Cl_2[/tex]) at elevated temperature.

Since we are balancing equations here, below is the balanced equation for Fe reacting with [tex]Cl_2[/tex] (offered at no extra charge):

[tex]2Fe + 3Cl_2[/tex] → [tex]2FeCl_3[/tex]

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Answer:

I think it's letter B

Explanation:

Hope it helps correct if I'm wrong

Answer:

[tex]\boxed {\boxed {\sf 155 \ g\ CO}}[/tex]

Explanation:

To convert from moles to mass, the molar mass must be used.

First, write the chemical formula for carbon monoxide. Since the carbon (C) comes first without a prefix, there is 1 carbon atom. The prefix mono- before oxide means 1, so there is also 1 oxygen (O) atom. The formula is CO.

Next, look up their molar masses on the Periodic Table.

Since there is 1 atom of each, the molar masses can be added.

Use this molar mass as a ratio.

[tex]\frac {28.01 \ g \ CO} {1 \ mol \ CO}[/tex]

Multiply by the given number of moles:5.55

[tex]5.55 \ mol \ CO *\frac {28.01 \ g \ CO} {1 \ mol \ CO}[/tex]

The moles of carbon monoxide cancel.

[tex]5.55 * \frac {28.01 \ g \ CO} {1 }[/tex]

Multiply.

[tex]155.4555 \ g \ CO[/tex]

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, it is the ones place. The 4 in the tenths place tells us to leave the 5.

[tex]155 \ g\ CO[/tex]

5.55 moles of carbon monoxide is about 155 grams.

Answer:

19.5

Explanation: using dalton law pt=p1+p2+p3...

the total pressure is 37.9 so to get pressure of gas b subtract pressure of gas a from total pressure.37.9-18.4 gas b equals 19.5

Answer:

6 volts

Explanation:

Use Ohm's Law:

V = IR

V = (2 amps)(3 ohms) = 6 volts

The caliper is a component which is used by a disc brake system for performing its function.

The disc braking system consist of components such as disc/rotor, a brake calliper and brake pads. When the brake pedal is pushed, brake fluid creates pressure that produces friction.

So we can conclude that the caliper is a component which is used by a disc brake system for performing its function.

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Answer:

49.1 g

Explanation:

First we convert 9.67x10²³ atoms of oxygen into moles, using Avogadro's number:

Then we calculate how many P₃O₉ moles there would be with 1.60 O moles:

Finally we convert 0.18 P₃O₉ moles into grams, using its molar mass:

Sulfate - SO4^-2

Sulfite - SO3^-2

Hydrogen Sulfate - HSO4-

Answer:

Hyba odpowiedzi c. Hyba

Mole can be defined as the number of molar mass units of the compound in the sample. The moles of ammonium dichromate in 325 grams sample is 1.29 mol.

The molar mass can be given as the mass of each atom in the formula unit. The molar mass of ammonium chromate is given as:

[tex]\rm (NH_4)_2Cr_2O_7=2(N)+4\;\times\;2(H)+2(Cr)+7(O)[/tex]

Substituting the mass of each atom to identify the molar mass of ammonium chromate:

[tex]\rm (NH_4)_2Cr_2O_7=2(14)+4\;\times\;2(1)+2(52)+7(16)\\ (NH_4)_2Cr_2O_7=28+8+104+112\\ (NH_4)_2Cr_2O_7=252 \;g/mol[/tex]

The molar mass of ammonium dichromate s 252 g/mol. The moles of the compound in 325 g is given as:

[tex]\rm Moles=\dfrac{mass}{molar\;mass} \\\\Moles=\dfrac{325\;g}{252\;g/mol}\\\\ Moles=1.29\;mol[/tex]

The moles of ammonium dichromate in the sample are 1.29 mol. Thus, option c is correct.

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Answer:

I guess this answer is c

yh

Answer:

b. (Actual yield ÷ theoretical yield) x 100

Explanation:

The percent yield will be a ratio of collected or actual yield to the theoretical yield. Like any percentage, you are determining a part of a whole. You multiply by 100 to convert your decimal answer from the division into a percentage.

Answer:

Has definite shape and volume

Solids have high melting point

Explanation:

Answer:

The gravitational potential energy and kinetic energy of this ball should be equal (assuming that there is no energy loss due to friction.)

Explanation:

The ball loses gravitational potential energy as it rolls down the hill. At the same time, the speed of the ball increases, such that the ball gains kinetic energy.

If there is no friction on this ball (and that the ball did not deshape,) all the gravitational potential energy that this ball lost would be converted to kinetic energy.

If the gravitational field strength [tex]g[/tex] is constant throughout, the gravitational potential energy of an object in that gravitational field would be proportional to its height.

If [tex]m[/tex] denote the mass of this ball, the gravitational potential energy ([tex]\rm GPE[/tex]) of this ball at height [tex]h[/tex] would be [tex]{\rm GPE} = (m \cdot g) \cdot h[/tex], which is proportional to [tex]h\![/tex].

The value of [tex]g[/tex] near the surface of the earth is indeed approximately constant (typically [tex]g \approx 9.8\; \rm m \cdot s^{-2}[/tex].)

At halfway between the top and bottom of this hill, the height of this ball would be [tex](1/2)[/tex] of its initial value (the value when the ball was at the top of the hill.) Because the [tex]\rm GPE[/tex] of this ball is proportional to its height, at halfway down the hill, the [tex]\rm GPE\![/tex] of this ball would also be [tex](1/2)\![/tex] its initial value.

However, if there was no friction on this ball (and that the ball did not deshape,) that [tex](1/2)[/tex] of the initial [tex]\rm GPE\![/tex] of this ball was not lost. Rather, these [tex](1/2)\![/tex] of the initial [tex]\rm GPE[/tex] would have been converted to the kinetic energy ([tex]\rm KE[/tex]) of this ball.

Hence, when the ball is halfway down the hill:

[tex]\displaystyle \text{GPE halfway down the hill} = \frac{1}{2}\, \text{Initial GPE}[/tex].

[tex]\begin{aligned}& \text{KE halfway down the hill}\\ &= \text{Initial GPE} - \text{GPE halfway down the hill}\\ &= \text{Initial GPE} - \frac{1}{2}\, \text{initial GPE}\\ &= \frac{1}{2}\, \text{Initial GPE}\end{aligned}[/tex].

Therefore:

[tex]\begin{aligned}& \text{GPE halfway down the hill} \\ &= \frac{1}{2}\, \text{Initial GPE} \\ &= \text{KE halfway down the hill}\end{aligned}[/tex].

In other words, under these assumptions, when this ball is halfway down the hill, the gravitational potential energy and the kinetic energy of this ball would be equal.

Answer:

The lithosphere solid part of the earth's crust

Answer:

the second one

Explanation:

metal doesn't act like a magnet because the domains don't have a preferred direction of alignment. Magnets are all aligned in a specific direction.

Answer:

%Cu = 66.4%

%S = 33.6%

Explanation:

Step 1: Given data

Step 2: Calculate the mass of S

The mass of CuS is equal to the sum of the masses of Cu and S.

mCuS = mCu + mS

mS = mCuS - mCu

mS = 9.56 g - 6.35 g = 3.21 g

Step 3: Calculate the percent by mass of each element

We will use the following expression.

%Element = mElement / mCompound × 100%

%Cu = mCu / mCuS × 100% = 6.35 g / 9.56 g × 100% = 66.4%

%S = mS / mCuS × 100% = 3.21 g / 9.56 g × 100% = 33.6%

Answer:

The correct answer is - [tex]\frac{P}{4}[/tex].

Explanation:

By the ideal gas equation is PV=nRT, the relation between pressure and volume can be drive if the temperature remains constant:

At constant temperature, for a fixed amount of gas, P∝

so the pressure is inversely proportional to volume. Therefore, if the volume is increased in this condition the pressure will decrease in the same manner and A decrease in the volume will lead to an increase in pressure.

Volume increased here = four times = 4V

pressure is P will decrease four times = P/4

Answer:

Increased evaporationof water vapour from sea or land will increase rainfall

Answer:

Explanation:

increased evaporation of water vapor from sea to land will increase rainfall

Solution :

It is given that :

Weight of the antacid tablet = 5.4630 g

4.3620 gram of antacid is crushed and is added to the stomach acid of 200 mL and is reacted.

25 mL of the stomach acid that is partially neutralized required 13.6 mL of NaOH to be titrated for a red end point.

27.7 mL of [tex]$NaOH$[/tex] solution is equivalent to [tex]$25 \ mL$[/tex] of the original stomach acid. Therefore, 13.6 mL of NaOH will take x [tex]$\frac{25\text{ mL of original stomach acid}}{\text{27.7 mL of NaOH}}$[/tex]

                                                             = 12.27 ml of the original stomach acid.

Answer:

32 mL

Explanation:

To answer this question we'll use Charles' law, which relates the temperature and volume of a gas at constant pressure:

V₁T₂=V₂T₁

Now we convert the given temperatures to K:

And given that V₁ = 45 mL, we can input the data:

And solve for V₂: