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2 years ago

Elements in group 2 are all called alkaline earth metals. What is most similar about the alkaline earth metals?

Chemistry

2 answers:

Eddi Din [679]2 years ago

6 0

Answer:

DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

Explanation:

german2 years ago

4 0

Answer:

Explanation: the last one

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A particular power source creates electricity through the use of giant magnets that are rotated to create a flow of electrons. T

Sloan [31]

From the electric generator to electric outlets in homes

6 0

2 years ago

Read 2 more answers

The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by

maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use Raoult’s Law to calculate the vapour pressure:

p₁ = χ₁p₁°

where

χ₁ = the mole fraction of the solvent

p₁ and p₁° are the vapour pressures of the solution and of the pure solvent

The formula for vapour pressure lowering Δp is

Δp = p₁° - p₁

Δp = p₁° - χ₁p₁° = p₁°(1 – χ₁) = χ₂p₁°

where χ₂ is the mole fraction of the solute.

Step 1. Calculate the mole fraction of glucose

n₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu

n₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O

χ₂ = n₂/(n₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62

Step 2. Calculate the vapour pressure lowering

Δp = χ₂p₁° = 0.018 62 × 23.8 torr = 0.4430 torr

Step 3. Calculate the vapour pressure

p₁ = p₁° - Δp = 23.8 torr – 0.4430 torr = 23.4 torr

3 0

3 years ago

Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec

Svetach [21]

The overall reaction is given by:

$Br_{2}(g) + 2 NO(g) \rightarrow 2 NOBr(g)$

The fast step reaction is given as:

$NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})$

The slow step reaction is given as:

$NOBr_{2}(g) + NO(g) \rightarrow 2 NOBr(g)$ (slow step $k_{2}$ )

Now, the expression for the rate of reaction of fast reaction is:

$r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]$

The expression for the rate of reaction of slow reaction is:

$r_{2}=k_{2}[NOBr_{2}] [NO]$

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as $[NOBr_{2}]$ takes place in this reaction.

The expression of rate of formation is:

$\frac{d(NOBr)}{dt}=r_{2}$

= $k_{2}[NOBr_{2}][NO]$ (1)

Now, consider that the fast step is always is in equilibrium. Therefore, $r_{1}=0$

$k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]$

$[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]$

Substitute the value of $[NOBr_{2}]$ in equation (1), we get:

$\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]$

= $k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]$

= $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$

Thus, rate law of formation of $NOBr$ in terms of reactants is given by $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$ .

4 0

3 years ago

The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?

Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

3 0

3 years ago

What principle chemistry

NNADVOKAT [17]

Is this a question ?

8 0

3 years ago