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2 years ago

7

Which factor explains why coal dust in an enclosed space is more explosive than coal dust blown outdoors into an open space ? A.

Temperature.
B.surface area
C.concentration
D.Volume
E.catalyst

2 answers:

ZanzabumX [31]2 years ago

8 0

The correct option is B.

Coal dust refers to the powered form of coal. Because of the high surface area of coal dust it is highly prone to dust explosion, which involves rapid combustion of fine particles that are suspended in the air; this usually occur in an enclosed place. Coal dust in an enclosed place is more explosive than coal dust that is blown outdoor in an open space because the coal dust in an enclosed place is more concentrated due to restricted space, thus it is more liable to explosion.

Amanda [17]2 years ago

5 0

C.concentration.much more explosive in a contained area.

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Selena is learning about how engineers choose metals for spacecraft. She discovers that spacecraft engineers often mix other met

Elena-2011 [213]

Ti is a transition metal.

<h3>Further explanation</h3>

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This is the atomic number of Titanium

The Atomic Number (Z) indicates the number of protons of an element

2. Ti

This is the symbol (chemical formula) of titanium

3. 47.90

This is the mass number of Titanium

Mass Number (A) = Number of protons + Number of Neutrons

Atomic Number (Z) = Mass Number (A) - Number of Neutrons

4. Titanium

This is the name of the element

To determine which group, alkaline, alkaline earth or transition it can be seen from the electron configuration

alkali and alkali earth : configuration from orbital 1s-7s

transition : configuration from orbital 3d-6d (Block d)

Electron configuration from Ti : [Ar] 3d²4s²

if we look at the configuration, then Ti is in the d block which is included in the transition metal

7 0

2 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

2 years ago

A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent

Degger [83]

Answer : The % of (+) limonene isomer = 79%

The % of (-) limonene isomer = 0%

The % of enantiomeric excess = 58%

Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.

Given,

% of pure limonene enantiomer = The % of (+) limonene isomer = 79%

Therefore, The % of (-) limonene isomer = 0%

Formula used :

$\%(+)\text{ isomer}=\frac{ee}{2}+50\%$

Where, ee → enantiomeric excess

Now, put all the values in above formula, we get the value of enantiomeric excess (ee).

${ee}=\frac{\%(+)-50\%}{2}$

$=\frac{79\%-50\%}{2}$

= 58%

7 0

1 year ago

Read 2 more answers

A chemist has 2.0 mol of methanol (CH3OH). The molar mass of methanol is 32.0 g/mol. What is the mass, in grams, of the sample?

rodikova [14]

Answer:

$\boxed {\boxed {\sf D. \ 64 \ grams }}$

Explanation:

Given the moles, we are asked to find the mass of a sample.

We know that the molar mass of methanol is 32.0 grams per mole. We can use this number as a fraction or ratio.

$\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}$

Multiply by the given number of moles, which is 2.0

$2.0 \ mol \ CH_3OH *\frac{32 \ g \ CH_3OH}{1 \ mol \ CH_3OH}$

The moles of methanol will cancel each other out.

$2.0 \ *\frac{32 \ g \ CH_3OH}{1 }$

The denominator of 1 can be ignored.

$2.0 * 32 \ g\ CH_3OH$

Multiply.

$64 \ g \ CH_3OH$

There are 64 grams of methanol in the sample.

3 0

1 year ago

Small beads of iridium-192 are sealed in a plastic tube and inserted through a needle into breast tumors. If an Ir-192 sample ha

Stells [14]

Answer:

296.1 day.

Explanation:

The decay of radioactive elements obeys first-order kinetics.

For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.

For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).

t is the time of the reaction (t = ??? day).

a is the initial concentration of Ir-192 (a = 560.0 dpm).

(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).

<em>∴ kt = lna/(a-x)</em>

(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).

(9.365 x 10⁻³ day⁻¹)(t) = 2.773.

<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>

5 0

2 years ago