Whos good at the periodic table cus im not help

larisa86 [58]

Additive color mixing involves multiple sources of light with different colors in each source. Subtractive color mixing involves a single source of light with different colors absorbing various wavelengths of the color spectrum. Secondary colors of one system serve as the primary colors for the other.

8 0

3 years ago

112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.

dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

$\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol$

For the given chemical reaction:

$2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)$

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = $\frac{12}{2}\times 0.778=4.668 mol$ of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

$\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol$

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

8 0

3 years ago

A 5.00 L sample of air at 0 C is warmed to 100.0 C. What is the new volume of the air? First, identify V1.

Paladinen [302]

Answer : The new volume of the air is, 6.83 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L$

Therefore, the new volume of the air is, 6.83 L

6 0

4 years ago

Nonmetals gain electrons under certain conditions to attain a noble gas electron configuration. How many electrons must be gaine

gogolik [260]

Answer:

1

Explanation:

For non metals to attain a noble gas configuration, they gain the number of electrons needed to attain the noble gas configuration of the noble gas at the end of their periods. This means that these non metals would only take up the configuration of the last element on their periods which of course is always a noble gas.

The last element on the hydrogen period or more conservatively the only other element on the hydrogen period is helium, with an atomic number of 2. The atomic number is the number of protons in he nucleus of an atom. For an electrically neutral atom, the number of electrons equal the number of protons.

Hence we can deduce that helium has 2 electrons while hydrogen has one electron. Thus for it to attain the configuration of helium, it just needs to gain one more electron

6 0

3 years ago

Thinking and questioning is the start of the scientific inquiry process. Please select the best answer from the choices provided

aev [14]

Yes thats true! You always have to think about the question or project before you start a science experiment! :)

8 0

3 years ago

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