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klio [65]
3 years ago
14

A 1-cubic-centimeter cube of sodium reacts more rapidly in water at 25°C than does a 1-cubic-centimeter cube of calcium at 25°C.

This difference in rate of reaction is most closely associated with the different
A) surface area of the metal cubes
B) nature of the metals
C) density of the metals
D) concentration of the metals
Chemistry
2 answers:
Tamiku [17]3 years ago
4 0

As given that the

a) temperature is same for both the metals

b) the volume of both the cubes is same (Same dimension)

The metals are sodium and calcium

Sodium has atomic number = 11

the electronic configuration is 1s2 2s2 2p6 3s1

It will attain noble gas configuration by easily losing one electron only

calcium has atomic number = 20

the electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s2

It will attain noble gas configuration by donating two electrons so calcium is less reactive than sodium

hence the difference in the rate of reaction is most closely associated with the different nature of the metals

ki77a [65]3 years ago
3 0
B) nature of the metals
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At a certain temperature the rate of this reaction is first order in HI with a rate constant of 0.0632 s
kirza4 [7]

Answer:

28.037\ \text{s}

Explanation:

[A]_0 = Initial concentration = 1.28 M

[A] = Final concentration = 0.17[A]_0

k = Rate constant = 0.0632 s

t = Time taken

For first order reaction we have the relation

kt=\ln\dfrac{[A]_0}{[A]}\\\Rightarrow t=\dfrac{\ln\dfrac{[A]_0}{[A]}}{k}\\\Rightarrow t=\dfrac{\ln\dfrac{[A]_0}{0.17[A]_0}}{0.0632}\\\Rightarrow t=28.037\ \text{s}

Time taken to reach the required concentration would be 28.037\ \text{s}.

8 0
3 years ago
What is the density (g/mL) of an object that has a mass of 0.03 kg and occupies a volume of 25 mL?
tatyana61 [14]
First convert the kg to g ----- 0.03kg = 30g
Then divide the mass by the volume ----- 30g ÷ 25mL = 1.2
The density is 1.2g/mL<span />
5 0
3 years ago
Read 2 more answers
Determine the partial negative charge on the bromine atom in a c−br bond. the bond length is 1.93 å and the bond dipole moment i
vaieri [72.5K]

Answer:

The value is x  =  0.151  \ e

Explanation:

From the question we are told that

The bond length is l  =  1.93\  \r  a =  1.93 *1 *10^{-10}  =1.93 *10^{-10}\  m

The bond dipole moment is \mu  = 1.40 d  = 1.40 *  3.33564 *10^{-30}  =  4.6699 *10^{-30} \  C \cdot m

Generally the dipole moment is mathematically represented as

\mu  =  Q *  l

Here Q is the partial negative charge on the bromine atom

So

Q =  \frac{\mu}{ l}

=> Q =  \frac{4.6699 *10^{-30}}{ 1.93 *10^{-10} }

=> Q = 2.42 *10^{-20} C

Generally

1 electronic charge(e) is equivalent to 1.60*10^{-19} C

So x electronic charge(e) is equivalent to Q = 2.42 *10^{-20} C

=> x  =  \frac{2.42 *10^{-20}}{1.60*10^{-19} }

=>     x  =  0.151  \ e

3 0
3 years ago
An object with a pre-weighed mass of exactly (and correctly) 0.54 g is given to 2 students. One student obtains a weight of 0.59
Alina [70]

Answer:

Both student have same percent error.

Explanation:

Given data:

Actual mass of object = 0.54 g

Measured value by one student = 0.59 g

Measured value by second student = 0.49 g

Percent error = ?

Solution:

Formula

Percent error = (measured value - actual value / actual value) × 100

Percent error of first student:

percent error = (0.59 g - 0.54 / 0.54 ) ×100

percent error =  9.3 %

Percent error of second student:

Percent error = (0.49 g - 0.54 / 0.54 ) ×100

Percent error = - 9.3 %

Both student have same percent error. The only difference is that first student measure the greater value then actual value and second student measure the less value then actual, however difference was same and gives same percent error.

3 0
3 years ago
Scientists Pam and Alejandro were discussing a liquid found in the science lab that was missing it's label. In classifying the u
strojnjashka [21]

Your answer for this would be A. Mixtures cannot be separated by physical meansA pure substance is heterogeneous.


Hope This Helps :)

6 0
3 years ago
Read 2 more answers
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