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3 years ago

Which of the following pairs of elements is most likely to form an ionic compound?

Chemistry

1 answer:

love history [14]3 years ago

4 0

Answer:

Option A = sodium and bromine

Explanation:

Sodium is metal while bromine is non metal. The bond formed between them is ionic.

Ionic bond:

Ionic bond is formed by the complete transfer of electron form metal to non metal.

Sodium is belongs to the group one while bromine is present in group seventeen. The number of valance electron of bromine are seven. Bromine require one electron to complete the octet, while sodium lose its one valance electron to gain noble gas stable electronic configuration. Thus sodium transfer its electron to bromine and gain positive charge and bromine gain its electron and form negative ion. The forces present between them are electrostatic forces which are very strong.

Na⁺Br⁻

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Baking powder is a mixture of baking soda and a mild edible acid such as tartaric acid

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3 years ago

How many moles of glucose can be produced

Arturiano [62]

Answer:

moles of glucose

2.3166 moles of glucose

Explanation:

The balance reaction for the formation of glucose is :

$6CO_{2}+6H_{2}O\rightarrow C_{6}H_{12}O_{6}+6O_{2}$

here , CO2 = carbon dioxide

H2O = water

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O2 = Oxygen

According to this equation :

6 mole of CO2 = 6 mole of H2O = 1 mole of C6H12O6 = 6 mole of O2

We are asked to calculate the mole of Glucose from carbon dioxide.

So,

6 mole of CO2 produce = 1 mole of C6H12O6

1 mole of CO2 will produce =

$\frac{1}{6}$ moles of glucose

13.9 moles of CO2 will produce :

$\frac{1}{6}\times 13.9$

=2.3166 moles of glucose

Note : first , Always calculate for one mole (By dividing)

. After this , multiply the answer with the moles given.

Always write the substance whose amount is asked(glucose) to the right hand side

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4 years ago

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

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