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3 years ago

What is an alkali metal with fewer than 10 protons in its nucleus?

Chemistry

1 answer:

Trava [24]3 years ago

6 0

Answer:

Lithium

Explanation:

Alkali metals are group of metals which are present in first group of periodic table. As we know atomic number is equal to number of protons contained by a particular element. Therefore, the alkali metals along with there number of protons are listed below;

Alkali Metal Number of Protons

Lithium 3

Sodium 11

Potassium 19

Rubidium 37

Cesium 55

Francium 87

Hence, it is cleared from above table that Lithium is having fewer protons than 10.

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What is the name of this compound? 2-butanone 2-pentanone 4-pentanone 4-butanone

UkoKoshka [18]

Answer:

2-pentanone

Explanation:

The oxygen atom is attached to the 2nd carbon atom in the longest linear chain, thus the prefix 2. the longest continuous carbon chain has five carbon atoms giving it the prefix pent-. the chemical is a ketone thus the suffix none.

7 0

4 years ago

Read 2 more answers

Write 266000 in scientific notation

kykrilka [37]

266000 = 2.66 x 10^5

hope this helps

8 0

4 years ago

Read 2 more answers

A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4

natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

$M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M$

8 0

3 years ago

Use Table 1: Analysis of Decay in the laboratory guide to answer this question: Scientists find a piece of wood that is thought

sashaice [31]

1/2=5750 years, 1/2(1/2)=1/4, (1/2)(1/2)(1/2)=1/8, (1/2)(1/2)(1/2)(1/2)=1/6
4 halflives have passed so 4(5750)=23000 years since the tree was chopped down
1000000 atoms (1/2)=500000 atoms(1/2)=250000(1/2)=125000(1/2)=62500 atoms would remain in the wood after 4 halflives
Dinosaurs became extinct around 62 million years ago, so if 14C's half life has a value of 5750 years, it would be gone or in such small amounts that dating would be ineffective today.
As Potassium decays into Argon in 1.3 billion years, apart from volcanic activity, it would enable geologists to effectively date things that are really, really, really old.

8 0

3 years ago

The molar volume of a certain solid is 142.0 cm3 mol−1 at 1.00 atm and 427.15 k, its melting temperature. the molar volume of th

kotykmax [81]

Answer:

$\Delta _{fus}H=3255.3J/mol$

$\Delta _{fus}S=7.62\frac{J}{mol*K}$

Explanation:

Hello,

Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:

$\frac{dp}{dT}=\frac{\Delta _{fus}H}{T\Delta _{fus}V}$

As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:

$p_2-p_1=\frac{\Delta _{fus}H}{\Delta _{fus}V}ln(\frac{T_2}{T_1} )$

Solving for the enthalpy of fusion we obtain:

$\Delta _{fus}H=\frac{(p_2-p_1)(V_2-V1)}{ln(\frac{T_2}{T_1})} =\frac{(11.84atm-1.00 atm)(156.6cm^3/mol-142.0cm^3/mol)}{ln(\frac{429.26K}{427.15K} )} \\\\\Delta _{fus}H=32127.3atm*cm^3/mol*\frac{101325Pa}{1atm}*(\frac{1m}{100cm} )^3\\\Delta _{fus}H=3255.3J/mol$

Finally the entropy of fusion is given by:

$\Delta _{fus}S=\frac{\Delta _{fus}H}{T_1} =\frac{3255.3J/mol}{427.15K}\\ \\\Delta _{fus}S=7.62\frac{J}{mol*K}$

Best regards.

5 0

3 years ago