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riadik2000 [5.3K]

3 years ago

15

The graph shows the changes in the phase of ice when it is heated.

1 answer:

mina [271]3 years ago

6 0

Answer:

100 degree celcius, because it is the melting point of ice ob

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What moon phase occurs 3-4 days after a waning gibbous

Artyom0805 [142]

Third quarter (or last quarter)

4 0

3 years ago

An atom that loses or gains electrons becomes a/an:

Ilia_Sergeevich [38]

Answer: Ion

Explanation:

4 0

3 years ago

Read 2 more answers

One mole of ice at 0°C is added to two moles of water at 50°C under a constant external pressure of 1 atm. This process is carri

kotykmax [81]

Answer:

T2 = 29.79°C

Explanation:

Equliibrium signifies that heat loss = heat gained

Heat gained by Ice;

H = ML

Mass, M = Number of moles * Molar mass = 1 * 18 = 18g

l = 6.01 k J m o l = 334 J/g

C = 4.186 J/g

H = 18(334)

H = 6012

Heat lost by water

H = MCΔT

H = 18 * 4.186 * (50 - T2)

H = 3767.4 - 75.348T2

Since H = H, we have;

6012 = 3767.4 - 75.348T2

- 75.348T2 = 3767 - 6012

T2 = 2245 / 75.348

T2 = 29.79°C

8 0

2 years ago

Spell out the full name of the compound.<br> Submit<br> Help plz

nekit [7.7K]

Explanation:

A=Butan-2-one

B=Pentan-3-one

3 0

3 years ago

Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon

Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

<u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

<u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

For Oxygen = $\frac{3.57}{3.57}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

Now, taking the ratio of carbon atoms in both the samples, we get:

$C_1:C_2=\frac{1}{2}:1=1:2$

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0

3 years ago