Which is least likely to occur after an experiment is conducted to a test a hypothesis?
The data is analyzed to see if it supports or refutes the hypothesis.
• The same experiment is conducted again to see if the data are reliable.
The hypothesis becomes a theory if the results support it.
O A new experiment is designed to provide additional data about the hypothesis.

Answer:

0.246 M

Explanation:

The reaction that takes place is:

First we calculate how many HCl moles reacted, using the given volume and concentration:

Then we convert HCl moles into Sr(OH)₂ moles, using the stoichiometric coefficients of the balanced reaction:

Finally we calculate the concentration of the original Sr(OH)₂ solution:

Answer:

B is correct :)

Explanation:

Trust me I just took the test

Answer:

824 g

Explanation:

First we convert 7.0 x 10²⁴ formula units of magnesium chloride (MgCl₂) into moles, using Avogadro's number:

There are two Cl moles per MgCl₂ moles, meaning that we have (2 * 11.62) 23.24 moles of Cl

Finally we convert 23.24 moles of chlorine into grams, using chlorine's molar mass:

Answer:



A substance, mostly liquid that donates a proton or accepts an electron pair in reactions. An acid increases the concentration of H+ ions. A base is a substance that releases hydroxide (OH-) ions in aqueous solution, donates electrons and accepts protons.

[tex]{\underline{\underline{\bf{\red{\:\:Acid\:\:}}}}}[/tex]

[tex]{\underline{\underline{\bf{\orange{\:\:Base\:\:}}}}}[/tex]

Answer:

CaCl2 + Na2S --> 2NaCl + CaS?​

Explanation:

CaCl2 + Na2S --> 2NaCl + CaS?​

Answer:

pH = 8.18

Explanation:

The weak base, X, reacts with HCl as follows:

X + HCl → HX⁺ + Cl⁻

Where 1 mole of X with 1 mole of HCl produce 1 mole of HX⁺ (The conjugate acid of the weak base).

Now, using H-H equation for bases:

pOH = pKb + log [XH⁺] / [X]

Where pOH is the pOH of the buffer (pH = 14 -pOH)

pKb is -log Kb = 5.824

And [X] [HX⁺] are the molar concentrations of each specie

Now, at the neutralization of the half of HX⁺, the other half is as X, that means:

[X] = [HX⁺]

And:

pOH = pKb + log [HX⁺] / [X]

pOH = 5.824 + log 1

pOH = 5.824

pH = 14-pOH

Answer:

[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]

Explanation:

From the question we are told that

Light wavelength  [tex]\lambda_l=655nm[/tex]

Light wavelength atom fall [tex]x_L=5th-2nd[/tex]

Photon wavelength  [tex]\lambda_p=435nm[/tex]

Photon wavelength atom fall [tex]x_P=^th-2nd[/tex]

Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by

[tex]\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )[/tex]

Therefore for initial drop of 5th to 2nd

[tex]\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )[/tex]

Therefore for initial drop of 6th to 2nd

[tex]\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]

Generally we subtract (5th to 2nd) from (6th to 2nd)

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )-\frac{1}R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{5^2}-\frac{1}{6^2} )[/tex]

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}[/tex]

[tex]\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}[/tex]

[tex]\frac{1}{\lambda_{5-6}}=1.33*10^{-3}[/tex]

Therefore for 6th to 5th stage is mathematically given by

[tex]\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}[/tex]

[tex]\frac{1}{\lambda_{6-5}}=751.879nm[/tex]

[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]

Answer:

b I hope this is right good luck

Answer:

The temperature of this sample changes by 1.052 degrees Celsius

Explanation:

As we know

[tex]Q = mc\Delta T[/tex]

Where m is the mass of the substance

c is the specific heat of the substance

and [tex]\Delta T[/tex] is the change in temperature

Substituting the given values in above equation, we get -

[tex]114.6 = 108.9 * 1 * \Delta T\\\Delta T = 1.052[/tex]degree Celsius

The temperature of this sample changes by 1.052 degrees Celsius

Answer:

Imine can be isolated from the reaction mixture as water is continuously removed from the reaction chamber

Explanation:

In this reaction, a non -aqueous solvent  is not used (not mentioned in the question). Thus, we can say that there is continuous removal water under suitable reacting conditions and hence the imine formed is left behind.

Answer:

When sound is created, the air particles vibrate and collide with each other, causing the vibrations to pass between air particles. The vibrating particles pass the sound through to a person's ear and vibrate the ear drum. Light travels much faster than sound through air.

Explanation:

Answer:

The empirical formula of C6 H12,

Answer: a. Cathode

b. Galvanic cell

c. Anode

d. Electrolytic cell

e. half reaction

Explanation:

Galvanic cell or Electrochemical cell is defined as a device which is used for the conversion of the chemical energy produced in a spontaneous redox reaction into the electrical energy.

Electrolytic cell is a device where electrical energy is used to drive a non spontaneous chemical reaction.

In the electrochemical cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.  Thus the electrons are produced at anode and travel towards cathode.

The balanced two-half reactions will be:

Oxidation half reaction : [tex]M\rightarrow M^{n+}+ne^-[/tex]

Reduction half reaction : [tex]N^{n+}+ne^-\rightarrow N[/tex]

Thus the overall reaction will be: [tex]M+N^{n+}\rghtarrow M^{n+}+N[/tex]

Answer:True

Explanation:

PLZ GIVE ME BRAINLIEST, I NEED IT

Answer:

B. Solidification

Hope this helped! :)

Answer:

I  think it A

Explanation:

Answer:

its d i think

Explanation:

Answer:

true

Explanation:

Answer:

1/16

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 269 years

Time (t) = 1076 years

Fraction remaining =?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 269 years

Time (t) = 1076 years

Number of half-lives (n) =?

n = t / t½

n = 1076 / 269

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the fraction of the original amount remaining. This can be obtained as follow:

Let N₀ be the original amount.

Let N be the amount remaining.

Number of half-lives (n) = 4

Fraction remaining (N/N₀ ) =?

N = 1/2ⁿ × N₀

N = 1/2⁴ × N₀

N = 1/16 × N₀

Divide both side by N₀

N/N₀ = 1/16

Thus, the fraction of the original amount remaining is 1/16

Answer:

Explanation:

We have the three equations:

[tex]NH_{3(g)} + HCl_{(g)} => NH_4Cl_{(s)} ..... \Delta H = ? (1)\\2HCl_{(g)} => H_{2(g)} + Cl_{2(g)} .... \Delta H = +184.6 kJ (2)\\2H_{2(g)} + 1/2N_{2(g)} + 1/2Cl_{2(g)} => NH_4Cl_{(s)} ..... \Delta H = -314.2 kJ (3)\\ N_{2(g)} + 3H_{2(g)} => 2NH_{3(g)} .... \Delta H = +184.6kJ (4)[/tex]

(can you double check that it is 184.6kJ for both equations 2 and 4 because it seems unlikely). We need to solve for equation 1 by addition and changing equations 2, 3 and 4. After possibly some trial and error, we can find that if we flip equations 4, multiply equation 3 by 2, add the equations together, and then finally divide by 2, we can get equation 1. We will get the answer of -314.2 kJ. However, I am again skeptical about the delta H values for equation 2 and 4 so double check that. This method might be super confusing and it is really hard to explain. So what I would suggest you to watch videos on Hess' law.

Answer:

Eutrophication

Explanation:

Eutrophication results when there is excess levels of phosphates and nitrates in its water of lake.

When there is high level of nutrition level in the water body of lake, then the productivity of water plants increases thereby leading to algal boom and hence Eutrophication.

Answer:

hello your question is incomplete attached below is the complete question

answer : To 2 significant digits = 5500 kg

Explanation:

Given

C(s) + O2 (g)  ----------> CO2  ( g )

mass of CO2 = 8.9 * 10^6 g

number of moles =  8.9 * 10^6 / 12  = 7.416 * 10 ^5  mol

same amount of moles is needed by O2 hence

attached below is the detailed  solution

Answer:

[tex]T_2=484.8K[/tex]

Explanation:

Hello there!

In this case, according to the the variable temperature and pressure and constant volume, it turns out possible for us to calculate the new temperature via the Gay-Lussac's law as shown below:

[tex]\frac{T_2}{P_2} =\frac{T_1}{P_1}[/tex]

Thus, by solving for the final temperature, T2, we obtain:

[tex]T_2 =\frac{T_1P_2}{P_1}[/tex]

So we plug in the given data to obtain:

[tex]T_2 =\frac{303K*200atm}{125atm}\\\\T_2=484.8K[/tex]

Best regards!

Explanation:

[tex]pH = - log[H {}^{ + } ] \\ = - log(5.6 \times {10}^{ - 4} ) \\ = 3.25[/tex]

Answer:

480.9K

Explanation:

V1=21L

V2=38.7L

T1=261K

T2=?

Using Charles law,

V1/T1=V2/T2

Inputting the given values,

21/261=38.7/T2

T2=480.9K