All categories

3 years ago

Something with a density greater than 1.00 g ml

1 answer:

5 0

Something with a density greater than 1.00 g ml will sink in waterThe density of water is 1 gm/cm^3=1000 kg/m^3, therefore, any element or object with density greater than 1 will sink in water.
Examples of elements with density greater than 1:
Aluminum (2700 kg/m^3 = 2.7 gm/cm^3)
Nickel ( 8910 kg/m^3 = 8.91 gm/cm^3)

You might be interested in

A weak acid, ha, has a p��a of 4. 524. If a solution of this acid has a ph of 4. 799 , what percentage of the acid is not io

insens350 [35]

Answer:

highly endothermic

Explanation:

6 0

2 years ago

What is the Molarity of a solution of HNO3 if it contains 12.6 moles in a 0.75 L solution? *

Korolek [52]

Answer:

M = 16.8 M

Explanation:

Data: HNO3

moles = 12.6 moles

solution volume = 0.75 L

Molarity is represented by the letter M and is defined as the amount of solute expressed in moles per liter of solution.

$M=\frac{moles}{solution volume}$

The data is replaced in the given equation:

$M=\frac{12.6 mol}{0.75L}=16.8\frac{mol}{L}$

7 0

3 years ago

Read 2 more answers

____Cl2+____H2=____HCl

Genrish500 [490]

Answer:

H2 + 2 Cl2 = 2 HCl2

6 0

3 years ago

What is the total volume of gaseous products formed when 116 liters of butane (C4H10) react completely according to the followin

Contact [7]

Answer: The total volume of the gaseous products is 1044.29 L

Explanation:

We are given:

Volume of butane = 116 L

At STP:

22.4 L of volume is occupied by 1 mole of a gas

So, 116 L of volume will be occupied by = $\frac{1}{22.4}\times 116=5.18mol$ of butane

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

For carbon dioxide:

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 5.18 moles of butane will produce = $\frac{8}{2}\times 5.18=20.72mol$ of carbon dioxide

Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L

For water vapor:

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water vapor

So, 5.18 moles of butane will produce = $\frac{10}{2}\times 5.18=25.9mol$ of water vapor

Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L

Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L

Hence, the total volume of the gaseous products is 1044.29 L

3 0

3 years ago

You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,

bonufazy [111]

Answer:

$\frac{[F^{-}]}{[HF]}$ is larger

Explanation:

$pK_{a}=-logK_{a}$ , where $K_{a}$ is the acid dissociation constant.

For a monoprotic acid e.g. HA, $K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$ and $\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}$

So, clearly, higher the $K_{a}$ value , lower will the the $pK_{a}$

In this mixture, at equilibrium, $[H^{+}]$ will be constant.

$K_{a}$ of HF is grater than $K_{a}$ of HCN

Hence, $(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})$

So, $\frac{[F^{-}]}{[HF]}$ is larger

5 0

3 years ago