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3 years ago

Please answer soon!!!!!

Chemistry

1 answer:

fgiga [73]3 years ago

5 0

Answer:

7.27 grams

Explanation:

Format= starting amount (g)·(1 mol/molar mass)·(amount of wanted/amount of starting)·(molar mass of wanted/1 mol)

$41.0 x \frac{1 mol}{17.031}x\frac{3 H2}{2NH3} x\frac{2.016}{1 mol} =7.27$

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Convert 32.56 km/hr into ft/hr

Gekata [30.6K]

Note that
1 m = 3.2808 ft

Therefore
1 km = 3280.8 ft
and
$32.56 \, \frac{km}{h} = (32.56 \, \frac{km}{h})*(3280.8 \, \frac{ft}{km}) = 1.0682 \, \times 10^{5} \, \frac{ft}{h}$

Answer: 1.0682 x 10⁵ ft/hr

8 0

2 years ago

A drop of water with a mass of 0.48 g is vaporized at 100 ∘C and condenses on the surface of a 55- g block of aluminum that is i

bagirrra123 [75]

The final temperature in Celsius of the metal block is 49°C.

<h3>How to find the number of moles ?</h3>

Moles water = $\frac{\text{Given mass}}{\text{Molar Mass}}$

= $\frac{0.48\ g}{18\ \text{g/mol}}$

= 0.0266 moles

Heat lost by water = 0.0266 mol x 44.0 kJ/mol

= 1.17 kJ

= 1170 J [1 kJ = 1000 J]

Heat lost = Heat gained

Heat gained by aluminum = 1170 J

1170 = 55 x 0.903 (T - 25) = 49.7 T - 1242

1170 + 1242 = 49.7 T

T = 48.5°C (49°C at two significant figures)

Thus from the above conclusion we can say that The final temperature in Celsius of the metal block is 49°C.

Learn more about the Moles here: brainly.com/question/15356425

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7 0

1 year 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

Read 2 more answers

What is the total number of valence electrons in the lewis structure of xeo4?

malfutka [58]

Here is a picture of which shows you how many valence electrons are in the Lewis structure of xeo4