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Lilit [14]
4 years ago
9

470,809 mi express in an integer

Chemistry
2 answers:
navik [9.2K]4 years ago
8 0

<u>Answer:</u> The given number is already an integer.

<u>Explanation:</u>

Integer is defined as a whole number which is not a fraction. This number can be positive, negative or zero.

Decimal numbers are not considered as integers. They are rounded off to write an integer.

<u>For Example:</u> 4, -2 and 0, all are considered as integers.

We are given a number having value 470,809 miles. The number is already a whole positive number.

Thus, the given number is already an integer.

cupoosta [38]4 years ago
7 0
-470,809 m is in an integer
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<h3>0.020 × 1000 × 100</h3>

<h2>= 2000 mg of Sn</h2>

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Mercury is a liquid meteal but it is not hard? how can we say it is a metal
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4 years ago
A certain reaction at equilibrium has more moles of gaseous products than of gaseous reactants.(b) Write a statement about the r
mojhsa [17]

To be equal as K_p , K_c needs to be divided by (RT)^{\Delta n}

<h3>Briefly explained</h3>

When the amount of gaseous reactant is greater than the amount of gaseous product, where n_{products} < n_{reactants},

then = K_p < K_c

It's because \Delta n is negative, which places (RT) in the denominator. This is how the equation will now appear.

K_p = K_c(RT)^{-\Delta n} = \frac{ K_c}{(RT)^{\Delta n}}

Here you can observe the value of K_c  is greater than K_p. To be equal as K_p , K_c needs to be divided by (RT)^{\Delta n}

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brainly.com/question/11336012

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6 0
2 years ago
Will the volume of a 1.0 L sample of gas at STP change if the pressure and temperature are both doubled
Dmitry_Shevchenko [17]

Answer:

The volume will not change.  This belongs in Ripley's Believe It or Not.

Explanation:

The combined gas law can be used to model both the initial (1) and ending (2) states of a gas when pressure (P), temperature (T) and/or volume (V) change, but the number of moles does not.  Remember that temperature must always be in Kelvin.

P1V1/T1 = P2T2/T2

Rearranging for V2:

V2 = V1(T2/T1)(P1/P2)

I've arranged the pressure and temperature terms as ratios.  This makes it easier to see what impact changes will have, plus the units conveniently cancel for both.

(V2) = (1 L)(T2/T1)(P1/P2)

We are told that P2 and T2 are both doubled:

   (T2/T1) = 2

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V2 = (1 L)(T2/T1)(P1/P2)

V2 = (1 L)(2)(1/2)

V2 = (1 L)(2)(1/2)

V2 - 1 L

The volume does not change.  Bummer.

7 0
2 years ago
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