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3 years ago

At STP, what physical property of aluminum always remains the same from sample to sample?

1 answer:

4 0

At STP, the <span>physical property of aluminum that always remains the same from sample to sample is density. The standard temperature is actually 273 degree kelvin and the standard pressure is 1 atmospheric pressure. I hope that this is the answer that has actually come to your desired help.</span>

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The speed of light in a vacuum is 2.998 x 108 m/s. What is its speed in (a) km/h; (b) mi/min?

love history [14]

Answer :

(a) The speed of light in km/hr is, $10.7928\times 10^8Km/hr$

(b) The speed of light in mi/min is, $1.117\times 10^7mi/min$

Explanation :

(a) We are given the speed of light is, $2.998\times 10^8m/s$

As we know that,

$1m=\frac{1}{1000}m$

and,

$1s=\frac{1}{3600}hr$

By using both the conversion, we get:

$2.998\times 10^8m/s$

$\Rightarrow \frac{2.998\times 10^8m}{1s}\times \frac{3600s}{1hr}\times \frac{1km}{1000m}$

$\Rightarrow 10.7928\times 10^8Km/hr$

The speed of light in km/hr is, $10.7928\times 10^8Km/hr$

(b) We are given the speed of light is, $2.998\times 10^8m/s$

As we know that,

$1m=\frac{1}{1609.34}mi$

and,

$1s=\frac{1}{60}min$

By using both the conversion, we get:

$2.998\times 10^8m/s$

$\Rightarrow \frac{2.998\times 10^8m}{1s}\times \frac{60s}{1min}\times \frac{1mi}{1609.34m}$

$\Rightarrow 1.117\times 10^7mi/min$

The speed of light in mi/min is, $1.117\times 10^7mi/min$

5 0

3 years ago

Helppppppp is due for tomorrow

Zielflug [23.3K]

The right answer is B

6 0

3 years ago

Read 2 more answers

In this reaction, the covalent bonds in the molecules of hydrogen and oxygen are broken.

Westkost [7]

Answer: Endothermic

Explanation:

Bond-breaking is an endothermic process. Energy is released when new bonds form. Bond-making is an exothermic process. Whether a reaction is endothermic or exothermic depends on the difference between the energy needed to break bonds and the energy released when new bonds form.

7 0

3 years ago

In this equation 2mg+o2=2mgo what is the coefficient to the oxygen molecule

Leya [2.2K]

Hello!

In this chemical equation, there are two main elements. Those elements are magnesium (Mg) and oxygen (O).

Molecules are a group of atoms bonded together, and in this case, the oxygen molecules is O2, but not 2MgO because there are two different molecules in this compound.

Coefficients are a number that multiply an entire molecule, and are also placed in front of it. In this case, O2 has no coefficient, but a subscript of 2. But, every molecule in this equation must have a coefficient and if add a coefficient of 1 to the oxygen molecule, it does nothing to the overall balancing of equations.

Therefore, the coefficient to the oxygen molecule is choice C, 1.

3 0

3 years ago

The standard heat of formation of a branched alkane is -35 kcal/mol; the standard heat of formation of the unbranched version of

AURORKA [14]

Answer:

-625 kcal/mol

Explanation:

The method to solve this question is based on Hess´s law of constant heat of summation which allows us to combine the enthalpies of individual reactions for which we know their enthalpy to obtain the enthalpy change for a desired reaction.

We are asked to calculate the standard enthalpy of formation of combustion of an unbranched alkane :

CnHn+2 unbranched + O₂ ⇒ CO₂ + H₂O ΔcHº = ?

where CnH2n+2 is the general formula for alkanes.

and we are given information for

n C+ (2n + n)/2 H₂ ⇒ CnHn+2 unbranched ΔfHº = -35 kcal/mol (1)

n C+ (2n + n)/2 H₂ ⇒ CnHn+2 branched ΔfHº = -28 kcal/mol (2)

CnHn+2 branched + O₂ ⇒ CO₂ + H₂O ΔcHº = -632 kcal/mol (3)

If we reverse (1) and add it to the sum (2) and (3) we get the desired equation for the combustion of the unbranched alkane:

CnHn+2 unbranched + O₂ ⇒ CO₂ + H₂O

Thus

ΔcHº unbranched = + 35 kcal/mol + (-28 kcal/mol) + (-632 kcal/mol)

= -625 kcal/mol

5 0

4 years ago