All categories

2 years ago

Atoms was determined to be 143 picometers. What is the percent error of this measured value?

Chemistry

1 answer:

jonny [76]2 years ago

8 0

Answer:

9.1% error

Explanation:

Full Question:

A measured value for the atomic radius of platinum atoms was determined to be 143 picometers. What is the percent error of this measured value?

You might be interested in

How many moles of ions are in 1.45 mol of K2SO4 ?

leonid [27]

Answer:

Explanation:

First we look generally at what makes K2SO4.

In one mole of K2SO4, there are 2 moles of the potassium ion (K+) and 1 mole of sulfate ion (SO4 2-).

Knowing that; in 1.75 moles of K2SO4, there must be 2 x 1.75moles of potassium ion (K+) and 1 x 1.75moles of Sulfate ion (SO4 2-)

This gives us 3.5moles of K+ and 1.75moles of SO4 2-

3 0

2 years ago

What is produced on the top of a pool of water by the mutual attraction between water molecules?

forsale [732]

The answer is surface tension

6 0

2 years ago

Which of the following accurately represents the relationship between ceramic and metal?

evablogger [386]

Search Results
Featured snippet from the web
The atoms in ceramic materials are held together by a chemical bond. The two most common chemical bonds for ceramic materials are covalent and ionic. For metals, the chemical bond is called the metallic bond. The bonding of atoms together is much stronger in covalent and ionic bonding than in metallic.

4 0

2 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

2 years ago

Write a half-reaction for the oxidation of the manganese in MnCO3(s) to MnO2(s) in neutral groundwater where the carbonate-conta

Leokris [45]

Answer:MnCO3+2H2O----->MnO2+ HCO3-+2e-+3H+

Explanation:The equation to be balanced is

MnCO3 ------> MnO2+HCO3-

The oxidation number of Mn changes from +2 in MnCO3 to +4 in MnO2

Therefore two electrons must be added to the right as shown below:

MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.

To balance the equation,add3H+on the right,to cancel out the charges.

MnCO3 --------> MnO2+HCO3-+2e-+3H+

Adding H2O to balance Hydrogen and Oxygen atoms:

MnCO3+2H2O ------->MnO2+HCO3-+2e-+3H+

3 0

2 years ago