A type of potential energy that is associated with objects that can be stretched or compressed is called ??

If after heating the reaction between magnesium and hydrochloric acid, the mass of the magnesium chloride and beaker is measured

Allushta [10]

Since the beaker was heated we can asume that only magnesium cloride is left in the beaker, therfore the difference between the beaker with magnesium chloride and the empty beaker give the mass of magnisium chloride:

$\text{mass MgCl}_2=massbeakerwithMgCl_2-mass\text{ emprty beaker =37.83 g - 33.95g = 3.8}8g$

7 0

1 year ago

Sound molecules can "stack up" in front of airplanes traveling at very high

11111nata11111 [884]

The answer is C a combustion’s reaction

5 0

3 years ago

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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]

Answer: The ratio of carbon in both the compounds is 1 : 2

Explanation:

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: $Cu_2O\text{ and }CuO$

For Sample 1:

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:

Step 1: 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 = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

Step 2: 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 = $\frac{4.55}{2.26}=2.01\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

For Sample 2:

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:

Step 1: 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 = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

Step 2: 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 = $\frac{3.57}{3.57}=1$

Step 3: 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

Calculate the mass of water produced when 7.26 g of butane reacts with excess oxygen

MaRussiya [10]

Answer:

11.3 g.

Explanation:

Hello there!

In this case, since the combustion of butane is:

$C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O$

Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:

$m_{H_2O}=7.26gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}}*\frac{5molH_2O}{1molC_4H_{10}} *\frac{18.02gH_2O}{1molH_2O}$

Therefore, the resulting mass of water is:

$m_{H_2O}=11.3gH_2O$

Best regards!

4 0

2 years ago

Canadian fans of Justin bieber and Drake .

KatRina [158]

Answer:

what?

Explanation:

6 0

3 years ago

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