Which type of front occurs when cold air and warm air are next to each other, but are at a standstill?

21<2x+3 help fast please

kolezko [41]

Answer:

Explanation:

x in (-oo:+oo)

2 < (1/2)*x-3 // - (1/2)*x-3

2-((1/2)*x)+3 < 0

(-1/2)*x+2+3 < 0

5-1/2*x < 0 // - 5

-1/2*x < -5 // : -1/2

x > -5/(-1/2)

x > 10

x in (10:+oo)

(10:+oo)

7 0

2 years ago

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Chemical reaction for acrylic fiber (7th grade chemistry) pllllzzzzzz help ASAP.

NNADVOKAT [17]

Answer: The chemical reaction for fiber is that it oxidizes.

6 0

3 years ago

A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas

Grace [21]

The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL

3 0

3 years ago

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What is the charge on every atom? why is this the charge

inn [45]

Protons, neutron, and elecrons

8 0

3 years ago

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution : Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 $Cm^{3}$

First, find the Density of given substance.

Formula used :

$Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}$

Now,put all the values in this formula, we get

$Density=\frac{46.9 g}{3.5 Cm^{3} }$ = 13.4 g/ $Cm^{3}$

So, we conclude that the density of given substance (13.4 g/ $Cm^{3}$ ) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ $Cm^{3}$ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0

3 years ago

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