Match all the terms that apply to each element. Some terms may apply to more than one element.

a. Main group element
b. halogen
c. noble gas
d. inner transition element
e. alkaline earth
f. Non-metal
g. Alkali metal
h. Metalloid
i. Metal

1. Titanium
2. Gold
3. Argon
4. Thallium

If the salt water is heated, the water turns into steam and the salt remains because the water has a lower boiling point than the salt.

The  following points can be considered:

The process involved when salt water is heated:

Therefore, the answer is water has a lower boiling point than salt.

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Answer:

0.6 ohm

Explanation:

voltage = 1.5 V

current = 2.5 A

resistence = ?

V = IR

1.5 = 2.5 * R

1.5 / 2.5 = r

0.6 = R

therefore resistence is 0.6 ohm

Answer:

27 g/mol of E

Explanation:

Note that percentage by mass= mass of each element present. So, since there is 90% of E, there is 90g of E present. By implication, there are 10g of H corresponding to 10%H. Note that there is 100g of EH3

1 moles of E corresponds to 90 g of E

Mole ratio of E: H= 1:3

Thus

Number of moles of H = 10g/ 1g/ mol = 10 moles of H

Since E contains 1/3 the number of moles of H

Number of moles of E = 1/3 × 10 = 3.33 moles of E

Molar mass of E= mass of E/ number of moles of E

Since mass of E = 90 g

Molar mass of E = 90g/3.33 moles

Molar mass of E = 27 g/mol

Answer:

moles CO₂ produced from combustion of 7.9 mole of C₅H₁₂ = 39.5 moles CO₂

Explanation:

              C₅H₁₂      +   8O₂      =>      6H₂O      +      5CO₂

Given:   7.9moles    excess            _____            ? moles

From Equation, 1 mole C₅H₁₂ =============>  5 moles CO₂

Given            7.9 mole C₅H₁₂  =============>         X

Solving for 'X' using ratio and proportion ...

1mole C₅H₁₂ / 7.9mole C₅H₁₂ =  5mole CO₂ / X

=> X = 5 mole CO₂ x  7.9mole C₅H₁₂ / 1mole C₅H₁₂  =  39.5 moles CO₂

≅ 40 moles CO₂ (2 sig. figs.)

Actual yield over theoretical yield, then multiply by 100

Answer:

50 CI₂ molecules

Explanation:

By looking at the stoichiometric coefficients, we can tell that if 2 atoms of potassium (K) react with chlorine gas (CI₂), 1 chlorine molecule would react.

With that in mind we can calculate how many CI₂ molecules would react with 100 K atoms:

Answer:

the correct answer is c

Explanation:

becuase i had the same question

Answer:

it needs two electrons in the first and eight to fill the second.

Explanation:

Because neon has two atomic shells, it needs two electrons in the first and eight to fill the second. Neon has a total of ten electrons which means two filled shells.

Answer:

Traduzca al ingles para que pueda responder a su pregunta. Te dare la reapuesta en los comentarios

Answer:

no equation

Explanation:

Answer:

Amount of water

The thermometer

Explanation

In an experiment, there is always a dependent variable and an independent variable. The independent variable is manipulated and its effect on the dependent variable is observed.

The control is that factor in the experiment that must remain constant so that effect of the independent variable on the dependent variable can be observed.

In this case, the independent variable is the amount of sodium chloride while the dependent variable is the temperature at which the solution boils.

The controls must be the amount of water which must be held constant and the same thermometer used to measure the temperature so that the effect of the amount of sodium chloride on the temperature of the solution can be studied.

The following are the controls in the experiment:

The beaker with 100 mL of water without any sodium chloride.

The temperature of the Bunsen burner.

The type of thermometer used.

The control(s) in the experiment are the beaker with 100 mL of water without any sodium chloride. This beaker is used to compare the boiling points of the other beakers, which have different amounts of sodium chloride added.

The control beaker ensures that any differences in boiling point are due to the amount of sodium chloride added, and not to other factors, such as the temperature of the Bunsen burner or the type of thermometer used.

The other factors that could affect the boiling point of water, such as the humidity of the air or the altitude, are kept constant in the experiment.

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Answer:

0.077 mole of ClF₃.

Explanation:

The balanced equation for the reaction is given below:

Cl₂ + 3F₂ —> 2ClF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of ClF₃.

Finally, we shall determine the number of mole of ClF₃ produced by the reaction of 0.115 mole of F₂. This can be obtained as follow:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of ClF₃.

Therefore, 0.115 mole of F₂ will react to to produce = (0.115 × 2)/3 = 0.077 mole of ClF₃.

Thus, 0.077 mole of ClF₃ was obtained from the reaction.

The question is incomplete, the complete question is;

Groups of the periodic table correspond to elements with a. the same color b. the same atomic number c. similar chemical properties d. similar numbers of neutrons

Answer:

similar chemical properties

Explanation:

In the periodic classification of elements, elements are divided into groups and periods. Elements in the same group of the periodic table have the same number of outermost electrons and share very similar chemical properties.

Elements in the same period have the same number of shells and the same maximum energy level of the outermost electron. Chemical properties carry markedly across a period.

Answer:

a. True

Explanation:

From the basic concepts of acids and bases, we know that when a base accepts a hydrogen ion (H⁺), it forms a conjugate acid which can accept again the H⁺ ion:

B⁻ + H₂O ⇆ BH + OH⁻

The stronger the base, the weaker the conjugate base. Thus, as more strength has a base, lesser strength will have the conjugate base (it will not be able to accept again the H⁺ ion). For example, when water (H₂O) loses its H⁺ , it forms the conjugate base OH⁻. So, OH⁻ is the stronger base that can exist in an aqueous solution.

H₂O ⇆ H⁺ + OH⁻

In fact, strong bases are hydroxides, such as NaOH or KOH.

Answer: The mass of [tex]H_2O[/tex] produced is 2.52 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of hexane = 1.72 g

Molar mass of hexane = 86.18 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of hexane}=\frac{1.72g}{86.18g/mol}=0.020mol[/tex]

Given mass of oxygen gas = 8.0 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of water}=\frac{8.0g}{32g/mol}=0.25mol[/tex]

The chemical equation for the combustion of hexane follows:

[tex]2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O[/tex]

By stoichiometry of the reaction:

If 2 moles of hexane reacts with 19 moles of oxygen gas

So, 0.020 moles of hexane will react with = [tex]\frac{19}{2}\times 0.020=0.19mol[/tex] of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hexane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of hexane produces 14 moles of [tex]H_2O[/tex]

So, 0.020 moles of hexane will produce = [tex]\frac{14}{2}\times 0.020=0.14mol[/tex] of [tex]H_2O[/tex]

We know, molar mass of [tex]H_2O[/tex] = 18 g/mol

Putting values in above equation, we get:

[tex]\text{Mass of }H_2O=(0.14mol\times 18g/mol)=2.52g[/tex]

Hence, the mass of [tex]H_2O[/tex] produced is 2.52 g

Answer:

Explanation:

First we convert 52.80 g of NaCl to moles, using its molar mass:

To determine the molality, we convert 150.0 g of water to kg:

As for the molarity, we determine the total mass of solution:

And calculate the volume, using the density:

Finally we calculate the molarity:

Answer:

V = 365.54 mL

Explanation:

Given that,

The density of chloroform, d = 1.48 g/mL

The mass of chloroform, m = 541 g

We need to find the volume of chloroform.

We know that,

Density = mass/volume

So,

[tex]V=\dfrac{m}{d}\\\\V=\dfrac{541\ g}{1.48\ g/mL}\\\\=365.54\ mL[/tex]

So, the volume of chloroform is 365.54 mL.

Answer:

Idont no men sorry idont understand

Explanation:

Sorry

Answer: The standard way to report this result is [tex]32.9\pm 0.3 mL[/tex]

Explanation:

The standard method of representing a result is:

[tex]\text{Calculated value }\pm \text{ Standard deviation}[/tex]

The reporting of a result is done in correct significant figures.

We are given:

Calculated value = 32.9 mL

Standard deviation = 0.3 mL

Rule of significant figures applied when numbers are added or subtracted:

The number having less number of significant figures after the decimal point will determine the number of significant figures in the final answer.

Number of significant figures after the decimal point = 1

Hence, the standard way to report this result is [tex]32.9\pm 0.3 mL[/tex]

The main consequence of exposure to the chemicals at Love Canal was:

allergies.

hair loss.

upper respiratory disease.

Explanation:

Knowing the pH, you know the concentration of protons:

−log[H+]=pH=3.7

[H+]=10−3.7 M

Now, since the weak (monoprotic) acid dissociates into its conjugate base and a proton, the mols of protons are equimolar with the mols of conjugate base---the protons came FROM the weak acid, so the conjugate base that forms must be equimolar with the protons given out to the solvent.

HA⇌A−+H+

Hence, [A−]=[H+] in the same solution volume. Using the equilibrium constant expression, we get:

Ka=[H+]2eq[HA]eq

Don't forget that the HA form of HA had given away protons, so the mols of protons given away to generate A− is subtracted from the mols of (protons in) HA.

=[H+]2eq[HA]i−[H+]eq

=(10−3.7M)20.02M−10−3.7M

Ka=2.0105×10−6 M

Answer:

On heating, the van der Waals dispersion forces existing then will easily break as it has a low boiling point and sublimates into gas. On heating iodine in the test tube, iodine evolves as violet fuming gas.

Explanation:

Answer:

The correct answer is - 5L.

Explanation:

From the ideal gas equation -

pv=nRT

p = pressure

v =volume

Here nRT is constant so

P would be inversely proportioned to v

So, p1/p2 = v2/v1

Putting values:

2/4(2) = v2/20 (p2 = 4 times of P1)

2/8 = v2/20

V2 = 5

Thus, the correct volume at new pressure would be - 5 L.

Answer: hello your question lacks some data attached below is the missing data

answer :

a) 3-methyl heptane

b) 2-methyl pentane

c) 2-methyl heptane

d) 2-methyl hexane

e) 3-methyl hexane

Explanation:

we will select the longest carbon chain as the branched alkane and name it

a) 3-methyl heptane ( first diagram )

b) 2-methyl pentane ( second diagram )

c) 2-methyl heptane ( third diagram )

d) 2-methyl hexane ( fourth diagram )

e) 3-methyl hexane ( fifth diagram )

Note : sixth diagram = first diagram

Answer:

0.375mL of solution of nalorphine must be injected

Explanation:

A solution of 0.40% (w/v) contains 0.40g of solute (In this case, nalorphine), in 100mL of solution. To obtain 1.5mg of nalorphine = 1.5x10⁻³g of nalorphine are needed:

1.5x10⁻³g * (100mL / 0.40g) =

Answer:

d,e

Explanation:

Answer:

Platinum is especially good for this because it is unreactive, and does not produce a color in the flame which will mask the presence of other metals.

Hope this answer is right!

Answer:

Hey mate, here is your answer

1. Platinum doesn't impart any color to the flame.

2. It is not oxidised under the high temperature of the flame from a bunsen burner.

3. It is almost chemically inert. Even at high temperatures, it remains unattacked by free radicals / acid radicals.

Therefore, platinum wire is crucial for a flame test. Also, a platinum wire should be thoroughly cleaned before using it for a new flame test.

A platinum wire is cleaned by dipping it into concentrated HNO3 and then placing it in the non luminous part of the bunsen flame. Otherwise, the perviously tested radicals will impart color to the flame, which may cause confusion.

Explanation:

Hope it helps you

Answer:

C) isoleucine

Explanation:

Isoleucine is among nine necessary amino acids in humans (found in dietary proteins). It has a variety of physiological activities, including aiding tissue repair, nitrogenous waste detoxification, immunological stimulation, and hormonal production promotion. When attached at a binding site, they are capable of providing a side chain thereby increasing the hydrophobicity at the binding site.

Answer: The mass of water produced is 507.6 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of glucose = 846 g

Molar mass of glucose = 180 g/mol

Plugging values in equation 1:

[tex]\text{Moles of glucose}=\frac{846g}{180g/mol}=4.7 mol[/tex]

The given chemical equation follows:

[tex]C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O+energy[/tex]

By the stoichiometry of the reaction:

If 1 mole of glucose produces 6 moles of water

So, 4.7 moles of glucose will produce = [tex]\frac{6}{1}\times 4.7=28.2mol[/tex] of water

Molar mass of water = 18 g/mol

Plugging values in equation 1:

[tex]\text{Mass of water}=(28.2mol\times 18g/mol)=507.6g[/tex]

Hence, the mass of water produced is 507.6 g

Answer:

Approximately [tex]1.30 \times 10^{-2}[/tex], assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let [tex]\rm HA[/tex] denote this acid.

[tex]\rm HA \rightleftharpoons H^{+} + A^{-}[/tex].

Initial concentration of [tex]\rm HA[/tex] without any dissociation:

[tex][{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}[/tex].

After [tex]12.5\%[/tex] of that was dissociated, the concentration of both [tex]\rm H^{+}[/tex] and [tex]\rm A^{-}[/tex] (conjugate base of this acid) would become:

[tex]12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}[/tex].

Concentration of [tex]\rm HA[/tex] in the solution after dissociation:

[tex](1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}[/tex].

Let [tex][{\rm HA}][/tex], [tex][{\rm H}^{+}][/tex], and [tex][{\rm A}^{-}][/tex] denote the concentration (in [tex]\rm mol \cdot L^{-1}[/tex] or [tex]\rm M[/tex]) of the corresponding species at equilibrium. Calculate the acid dissociation constant [tex]K_{\rm a}[/tex] for [tex]\rm HA[/tex], under the assumption that this acid is monoprotic:

[tex]\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}[/tex].