HELP MEEEE! ‘’’’’’’’’’

Answer:

0.275

Explanation:

Given that:

Initial number of miles of radioactive isotope = 4.4 moles

How much if the sample remains after 4 half life's

Half-life refers to the time taken by a sample to reduce to half of it's initial value.

After the first half life :

Initial / 2 = 4.4 / 2 = 2.2 moles

Second half life :

2.2 / 2 = 1.1 moles

Third half life :

1.1 / 2 = 0.55 Moles

Fourth half life :

0.55 / 2 = 0.275 moles

Answer:

Explanation:

The density of a substance can be found by using the formula

[tex]Density = \frac{mass}{volume} [/tex]

From the question

mass of CCl4 = 3.16 g

volume of CCl4 = 2.0 mL

Substitute the values into the above formula and solve for the Density

That's

[tex]Density = \frac{3.16}{2} [/tex]

We have the final answer as

Hope this helps you

Answer:

-891kj

Explanation:

We have the chemical equation as:

2SO2(g) + O2(g) ----2SO3; ∆H° = 198kj

If sulphur trioxide decomposes we have opposite reaction. Enthalpy of decomposition will not be positive for forward reaction.

2 moles of SO3 ----∆H° = -198Kj

1 mole = -198/2

9 moles = -198/2 x 9

= -891kj

Therefore change in enthalpy when 9 moles is decomposes is -891kj.

Explanation:

a chemical nomenclature is a set of rules to generate systematic names for chemical compounds.

Answer:

The weak base is hydroxylamine

Explanation:

The general equilibrium of a weak base (A), is:

A(aq) + H₂O(l) ⇄ HA⁺ + OH⁻

Where Kb is:

Kb = [HA⁺] [OH⁻] / [A]

As both HA⁺ and OH⁻ comes from the equilibrium of A; HA⁺ = OH⁻. So, you can take:

Kb = [OH⁻]² / [A]

OH⁻ is determined from pH, thus:

pOH = -log [OH⁻]

pOH = 14 - pH

pOH = 14 - 9.44

pOH = 4.56

[OH⁻] = 10^{-4.56}

[OH⁻] = 2.75x10⁻⁵

As the concentration of the weak base is 0.0678M:

Kb = [OH⁻]² / [A]

Kb = [2.75x10⁻⁵]² / [0.0678M]

Kb = 1.1x10⁻⁸ = Kb of Hydroxylamine.

Answer:

12 this is your answer

hope you like it

Answer:

66.441

I hope this helps!

Answer:

66.441

Explanation:

(5.391 × 10-7) + (2.5531 × 10-6)

(46.91) + (19.531)

66.441

Answer:

Explanation:

As the 50mL of water are in a big temperature will give heat to the other sample of water.

With the equation:

Q = m*C*T

Where Q is heat, m  is mass of the sample, C is specific heat of the sample (4.184J/molK for pure water) and T is temperature in K

We can determine the energy of each sample of water:

Sample 1:

Q = 30.0g*4.184J/molK*282K

Mass is 30.0g because density of water is 1g/mL

Q = 35396.6J

Sample 2:

Q = 50.0g*4.184J/molK*328K

Q = 68617.6J

Total energy is:

Q = 104014.2J

As total mass is 30+50 = 80g, final temperature is:

104014.2J = 80.0g*4.184J*T

310.7K is final temperature of the mixture

Answer:Luster

Explanation: i'm tired so it is luster because splotchy wouldn't be referring to the others so it would be referring to Luster the reflection of light of the object. Hope it helps

PLZ BRAINLILEST I ONLY NEED ONE MORE

Answer:

the answer is color

Explanation:

i got it right on my assignment lol

Answer:

42.857 seconds

Explanation:

Given that:

The average adult heart pumps about 84./mLs  of blood at 72 beats per minute.

The total volume = 3600 mL

The objective is to calculate how long it would take the average heart circulate 3600 mL of blood.

Since 84 mL of blood takes 1 seconds to pump,

assuming q is the time required for heart to pump 3600 mL of blood

[tex]Time \ ( q )= \dfrac{3600 mL \times 1\ seconds}{84 \ mL}[/tex]

TIme (q) = 42.857 seconds

Answer:  85.7 mL

Explanation:

Given the information from the question as plotted in the graph i will be uploading along side this answer,

Average of total volume of DCPIP used is

= (1.21 + 1.11 + 1.06)mL / 3

= 1.12 mL

and corresponding ( ascorbic acid ) is 0.70 g/L

Two parameter given as volume of DCPIP in final syringe and total volume of DCPIP are quite ambiguous

700mg ⇒ 1 L

THEREFORE volume that contains 60mg =  (1000/700) × 60 = 85.7 mL

Answer:

HELLO YOUR QUESTION IS INCOMPLETE ATTACHED BELOW IS THE COMPLETE QUESTION

answer : 0.033 mol  ( A )

Explanation:

To determine the number of moles of AI3 that will be reduced when 0.1 Faraday passes through a cell during the production of AI

AI3+  + 3 e -------- AI  this means that it will take 3 mol to reduce AI3 to AI

1 mol AI = 3 F  electrons

therefore : 1 F electrons  = 1/3 mol of AI

therefore : 0.1 F ( Faraday ) = 0.1 * 1/3  = 0.033 mol of AI is needed

Answer:

Tracy was able to remember what her teacher told her to do and she decided to try it,which was the right thing.

But since it's a demonstration then Tracy is supposed to wear all her protective equipments and with a teacher's permission.I hope this helps.Thanks

Answer:

The correct answer is 6.65 mg.

Explanation:

Based on the given information, the number of days given is 45.6 days. The original mass of sodium phosphate give is 175 mg. The half-life of phosphorus-32 is 14.3 days, therefore, the n or the number of half life will be,

n = 45.6/14.3 = 3.19

So, after 45.6 days, the mass of sodium phosphate sample left will be,

= Original mass × 1/2ⁿ

= 175 mg × 1/2 ^3.19

= 175 mg × 0.1096

= 34.3 mg of sodium phosphate left after 45.6 days

The molecular mass of Na₃³²PO₄ is 3 (23) + 32 + 4 (16) amu = 165 amu

Therefore, 165 grams of sodium phosphate comprise 32 grams of phosphorus.

So, 34.3 mg or 0.343 gram of sodium phosphate will contain,

= 0.343 × 32/165

= 6.65 mg of P³².

Answer:

The speed is 7 every second.

Equation:

14/2 = 7

A student graphed distance versus time for an object that moves 14 every 2 s. The speed of the object is 7 m/sec.

The speed is the distance traveled by an object with a specific unit of time. Speed is calculated by dividing distance by time. The SI unit of speed is a meter per second.

S = d / t

Distance is the space or extent between two things. It is measured in meters and kilometers.

Time is a variable that equals the distance divided by time.

The slope of a graph is the rate at which the y and x axis change together, the "rise over the run." Slope can be simply calculated using this equation:

Given, that distance = 14 meters

The time is 2 sec.

By the formula

d = 14 / 2 = 7 m /sec.

Thus, the speed of the object is 7 m/sec.

To learn more about speed, refer to the link:

https://brainly.com/question/7359669

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Answer:

Valence of iron in FeS2

Let the valency of iron be x

x+2(oxidation number of Sulphur)=0 as their is no charge on FeS2

As sulphur belongs to Group VI (chalcogens) so the valency/oxidation number of Sulphur is -2

x+2(-2)=0

x-4=0

x=4

So the valency/oxidation number of iron in FeS2 is +4

Explanation:

Hope it will help you monkey ^_^ it's crab

Answer:

Hope you are better now your parents must be worried about you get well soon  drink plenty of milk it will make your bones strong.

NOW TIME FOR WISHES ^_^

Answer:

a nonspontanenous reaction is forced to occur

Hope this helps :)

Answer:

The mass of FeSO₄.7H₂O  needed  [tex]\simeq[/tex] 225.89 grams

Explanation:

From the given question

The molecular weight of iron(ii) sulfate heptahydrate FeSO₄.7H₂O = 278.02 g/mol

Molarity = 3.25

Volume = 250.0 mL

The mass of FeSO₄.7H₂O  needed = [tex]\dfrac{moles \times molecular weight \times 250\ mL}{1000 \ mL}[/tex]

The mass of FeSO₄.7H₂O  needed  = [tex]\dfrac{3.25 \times 278.02 \times 250}{1000}[/tex]

The mass of FeSO₄.7H₂O  needed  = 225.89125 grams

The mass of FeSO₄.7H₂O  needed  [tex]\simeq[/tex] 225.89 grams

Answer:tell the instructor of the lab

Explanation:

Answer:

4.07 × 10⁹ mL

Explanation:

Step 1: Given data

Length of the pool (L): 60.0 m

Width of the pool (W): 35.0 m

Depth of water in the pool (D): 6.35 ft

Step 2: Convert "D" to m

We will use the relationship 1 m = 3.28 ft

[tex]6.35ft \times \frac{1m}{3.28ft} = 1.94m[/tex]

Step 3: Calculate the volume of water (V)

We will use the following expression.

[tex]V = L \times W \times D = 60.0m \times 35.0m \times 1.94 m = 4.07 \times 10^{3} m^{3}[/tex]

Step 4: Convert "V" to mL

We will use the relationship 1 m³ = 10⁶ mL.

[tex]4.07 \times 10^{3} m^{3} \times \frac{10^{6}mL }{1m^{3} } = 4.07 \times 10^{9} mL[/tex]

Answer:

Explanation:

The density of a substance can be found by using the formula

To calculate the Density in g/mL we must first convert the mass from kg to g and the volume from L to mL

For the mass

That's

1.34 kg

1kg = 1000 g

1.34 kg = 1000 × 1.34 = 1340 g

For the volume

That's

2.0 L

1 L = 1000 mL

2.0 L = 2 × 1000 = 2000 mL

So we have

mass = 1340 g

volume = 2000 mL

Substitute the values into the above formula and solve for the Density

That's

We have the final answer as

Hope this helps you

Answer:

Explanation:

Volume of air = 2.5 x 5.5 x 3.00

= 41.25 m³.

Density of air = 1.29 g / dm³

a dm³ = ( 0.1 )³m³

= .001 m³

density in m³ = 1.29 / .001 gm⁻³

= 1290 g / m³

mass of air in the room

= 41.25 x 1290 g

= 53212.5 g

= 53.2125 kg .

Explanation:

please mark me as brainlest

Answer:

C

Explanation:

Answer:

4 sig fis so look at the next number which is "4"

answer is 19.97

all numbers not zeros are significant

Answer:

The answer is on the screenshot

Explanation:

Answer:

Two (2) types of ions

Explanation:

Metal ions (cations) and negatively charged ions (anions) of either chloride, carbonate, sulphides or hydroxides are present during the selective precipitation.

Majorly, the process determines the metal ions present by initiating precipitation as certain conditions.

I hope this was helpful.

Answer:

0.15 (M) of NaCl            

Explanation:

An isotonic solution is defined as the solution having the same osmolarity, or the same solute concentration, as the another solution. If the two solutions are to be separated by any semipermeable membrane, then water will start flowing in equal parts from each solution and also into the other.  

In the context, since it is given that the red blood cells exerts the same osmotic pressure as a 0.15 (M) of NaCl solution.

Thus, 0.15 (M) of NaCl is a isotonic solution.  

Answer:

I ratio is

47 : 43 ratio

Answer:

(c) Nitric acid [tex]\rm HNO_3\, (aq)[/tex] is the limiting reactant.

(d) Approximately [tex]6.6\; \rm g[/tex] of [tex]\rm CuO\, (s)[/tex] will be in excess.

(e) The percentage yield of [tex]\rm Cu(NO_3)_2[/tex] is approximately [tex]91\%[/tex]. (Rounded to two significant figures, as in other quantities in the question.)

Explanation:

Start with the balanced chemical equation:

[tex]\rm CuO\, (s) + 2\; HNO_3\, (aq) \to Cu(NO_3)_2\, (aq) + H_2O\, (l)[/tex].

Look up relevant relative atomic mass data on a modern periodic table:

Calculate the formula mass of the species:

There are two reactants in this reaction: [tex]\rm CuO[/tex] and [tex]\rm HNO_3\, (aq)[/tex]. Assume that [tex]\rm CuO\![/tex] is the limiting one. In other words, assume that all the [tex]\rm CuO\!\![/tex] is consumed before [tex]\rm HNO_3\, (aq)\![/tex] was.

Consider: how many moles of [tex]\rm HNO_3\, (aq)\!\![/tex] would be required to convert all that [tex]7.0\; \rm g[/tex] of [tex]\rm CuO\!\!\![/tex] to [tex]\rm Cu(NO_3)_2[/tex]?

Calculate the number of moles of [tex]\rm CuO[/tex] formula units in [tex]7.0\;\rm g[/tex] of

[tex]\displaystyle n({\rm CuO}) = \frac{m}{M} = \frac{7.0\; \rm g}{79.545\; \rm g \cdot mol^{-1}} \approx 0.088001\; \rm mol[/tex].

Note the ratio between the coefficients of [tex]\rm CuO\, (s)[/tex] and [tex]\rm HNO_3\, (aq)[/tex]:

[tex]\displaystyle \frac{n(\mathrm{HNO_3\, (aq)})}{n(\mathrm{CuO\, (s)})} = \frac{2}{1} = 2[/tex].

Therefore:

[tex]\begin{aligned}&n(\text{$\mathrm{HNO_3\, (aq)}$, consumed (under assumption)})\\ &= \frac{n(\text{$\mathrm{HNO_3\, (aq)}$, consumed})}{n(\mathrm{CuO\, (s)})}\cdot n(\mathrm{CuO\, (s)}) \\ &\approx 2 \times 0.088001\; \rm mol \approx 0.17600\; \rm mol\end{aligned}[/tex].

On the other hand, how many moles of [tex]\rm HNO_3\, (aq)[/tex] are actually available?

Convert the volume of that [tex]\rm HNO_3\, (aq)[/tex] solution to the standard unit (liter.)

[tex]V(\rm HNO_3\, (aq)) = 50\; \rm mL = 0.050\; \rm L[/tex].

Calculate the number of moles of [tex]\rm HNO_3\, (aq)[/tex] in that [tex]0.20\; \rm M[/tex] solution:

[tex]n({\rm HNO_3\, (aq)}) = c\cdot V = 0.050\; \rm L \times 0.20\; \rm mol \cdot L^{-1} = 0.010\; \rm mol[/tex].

Apparently, the quantity of [tex]\rm HNO_3\, (aq)[/tex] required exceeded the quantity that is available. Therefore, the assumption is invalid, and [tex]\rm CuO[/tex] cannot be the limiting reactant. At the same time,

Since it is now known that all that [tex]0.010\; \rm mol[/tex] of [tex]\rm HNO_3\, (aq)[/tex] will be consumed, apply that coefficient ratio again to obtain the quantity of [tex]\rm CuO[/tex] consumed in this reaction:

[tex]\begin{aligned}&n(\text{$\mathrm{CuO}$, consumed})\\ &= n(\text{$\mathrm{HNO_3\, (aq)}$, consumed}) \left/\frac{n(\text{$\mathrm{HNO_3\, (aq)}$, consumed})}{n(\mathrm{CuO\, (s)})}\right. \\ &\approx (0.010 \; \rm mol) / 2 \approx 0.0050\; \rm mol\end{aligned}[/tex].

It was already shown that the formula mass of [tex]\rm CuO[/tex] is (approximately) [tex]79.545\; \rm g \cdot mol^{-1}[/tex]. Therefore, the mass of that [tex]0.0050\; \rm mol[/tex] formula units of [tex]\rm CuO\![/tex] would be:

[tex]\begin{aligned}& m(\text{$\rm CuO$, consumed}) \\&= n \cdot M \approx 0.0050\; \rm mol \times 79.545\; \rm g\cdot mol^{-1} \\&\approx 0.39773\; \rm g\end{aligned}[/tex].

Before the reaction, [tex]7.0\; \rm g[/tex] of [tex]\rm CuO[/tex] is available. Therefore, all that [tex]7.0\; \rm g - 0.39773\; \rm g \approx 6.6\; \rm g[/tex] of [tex]\rm CuO\![/tex] would be in excess.

Similarly:

[tex]\displaystyle \frac{n(\mathrm{HNO_3\, (aq)})}{n(\mathrm{Cu(NO_3)_2\, (aq}))} = \frac{2}{1} = 2[/tex].

Apply this ratio to find the theoretical yield of [tex]\rm Cu(NO_3)_2[/tex]:

[tex]n(\text{$\mathrm{Cu(NO_3)_2\, (aq)}$, theoretical yield}) \approx (0.010 \; \rm mol) / 2 \approx 0.0050\; \rm mol[/tex].

Find the mass of that [tex]0.0050\; \rm mol[/tex] of [tex]\rm Cu(NO_3)_2[/tex] formula units using its formula mass:

[tex]m(\text{$\rm Cu(NO_3)_2\, (aq)$, theoretical yield}) \approx 0.93777\; \rm g[/tex].

Calculate the percentage yield given that the actual yield is [tex]0.85\; \rm g[/tex]:

[tex]\begin{aligned}&\text{Percentage Yield} \\ &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\% \\ &= \frac{0.85\; \rm g}{0.93777\; \rm g} \times 100\% \approx 0.91\% \end{aligned}[/tex].

(Rounded to two significant figures.)

(c) The limiting reactant is HNO₃

(d) The mass of the excess reactant after the reaction is approximately 6.6 grams

(e) The percentage yield of copper (II) nitrate from the reaction is approximately 90.64%

The reason the above values are the correct values is as follows;

The given parameters are;

The mass of copper(II)oxide in the reaction = 7.0 g

The volume of the 0.20 M nitric acid, HNO₃ = 50 mL

(c) Concentration of the reactants

The molar mass of CuO = 79.545 g/mol

Number of moles = Mass/(Molar mass)

The number of moles of CuO = (7 g)/79.545 g/mol ≈ 0.088 moles

50 mL of 0.20 M HNO₃, contains 50/1000 × 0.2 = 0.01 moles of HNO₃

The chemical equation for the reaction is CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O

Therefore;

One mole of CuO reacts with two moles of HNO₃ to produce one mole of Cu(NO₃)₂ and one mole of H₂O

Therefore, 0.088 moles of CuO reacts with 2 × 0.088 = 0.176 moles of HNO₃

Given that there is only 0.01 moles of HNO₃, the limiting reactant is the HNO₃, which is not enough to completely react with the CuO which is the excess reactant

(d) The mass of the CuO that reacts with the 0.01 moles of HNO₃ is given as follows;

1 mole of CuO reacts with 2 moles HNO₃

0.01 moles of HNO₃ will react with 0.01/2 = 0.005 moles of CuO

Mass = Number of moles × Molar mass

The mass of 0.005 moles of CuO = 0.005 moles × 79.545 g/mol = 0.397725 grams

The mass of the CuO left = Initial mass - Reacting mass

∴ The mass of the CuO left = 7 grams - 0.397725 grams = 6.602275 grams

The mass of the excess reactant (CuO) after the reaction ≈ 6.6 grams

(e) The theoretical number of moles of copper (II) nitrate, Cu(NO₃)₂ produced = Half the number of moles of HNO₃ in the reaction

The number of moles of HNO₃ in the reaction = 0.01 moles

∴ The theoretical number of moles of copper (II) nitrate, Cu(NO₃)₂ produced = (1/2) × 0.01 moles = 0.005 moles

The molar mass of Cu(NO₃)₂ = 187.56 g/mol

The theoretical mass of Cu(NO₃)₂ produced = 0.005 moles × 187.56 g/mol = 0.9378 grams

The actual yield of copper (II) nitrate is 0.84 g

[tex]Percentage \ yield = \mathbf{\dfrac{Actual \ yield}{Theoretical \ yield} \times 100}[/tex]

Therefore;

[tex]\% \ yield \ of \ Cu(NO_3)_2 = \dfrac{0.85 \, g}{0.9378 \, g} \times 100 \approx 90.64 \%[/tex]

The percentage yield of copper (II) nitrate, %yield ≈ 90.64%

Learn more about percentage yield from chemical reactions here:

https://brainly.com/question/15443303