Ask question

Ask question

All categories

2 years ago

11

A sample of calcium oxide (CaO) has a mass of 2.80 g. The molar mass of CaO is 56.08 g/mol. How many moles of CaO does this samp

le contain?
0.0499
20.0
58.9
157

2 answers:

Olenka [21]2 years ago

4 0

Answer:

A) 0.0499

Explanation:

taurus [48]2 years ago

3 0

Answer: 0.0499

Explanation:

Got it from quizlet and it was correct on my test.

You might be interested in

An ate or ite at the end of a compound name usually indicates that the compound contains___________.

Ymorist [56]

Answer:

D

Explanation:

According to IUPAC nomenclature, a negative polyatomic ion has its name ending in 'ate' or 'ite'

As in carbonate (CO3^2-)

Arsenite (AsO3-)

3 0

3 years ago

Read 2 more answers

An ionic bond is a bond between

OleMash [197]

between two oppositely charged ions

7 0

3 years ago

Read 2 more answers

BRAINLIEST ASAP!!! Help

Andru [333]

Answer:

N2C14

Explanation:

<em> determined the bond type by looking if it is a metal or nometal</em>

<em>Ionic Bond:NM+M</em>

<em>Covalent Bond:NM+NM</em>

7 0

3 years ago

10. At 573K, NO2(g) decomposes forming NO and O2. The decomposition reaction is second order in NO2 with a rate constant of 1.1

leva [86]

Answer:

48.67 seconds

Explanation:

From;

1/[A] = kt + 1/[A]o

[A] = concentration at time t

t= time taken

k= rate constant

[A]o = initial concentration

Since [A] =[A]o - 0.75[A]o

[A] = 0.056 M - 0.042 M

[A] = 0.014 M

1/0.014 = (1.1t) + 1/0.056

71.4 - 17.86 = 1.1t

53.54 = 1.1t

t= 53.54/1.1

t= 48.67 seconds

Hence,it takes 48.67 seconds to decompose.

6 0

2 years ago

Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a

natali 33 [55]

Answer: Equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

$Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)$

We know that,

K = $K_{f} \times K_{sp}$

We are given that, $K_{f} = 1.0 \times 10^{31}$

and, $K_{sp} = 2.8 \times 10^{-16}$

Hence, we will calculate the value of K as follows.

K = $K_{f} \times K_{sp}$

K = $(1.0 \times 10^{31}) \times (2.8 \times 10^{-16})$

= $2.8 \times 10^{15}$

Thus, we can conclude that equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

4 0

3 years ago