All categories

2 years ago

30 pts for all answers pls

Chemistry

2 answers:

m_a_m_a [10]2 years ago

7 0

6.) 12600
7.) 0.57
8.) 3.10
9.) 365.1
10.) 5000

hope this helps

Oksanka [162]2 years ago

5 0

Explanation:

dont know just found it im in fifth grade so....... this is for the first one thats all i could do kk

You might be interested in

The solid compound, K2SO4, contains?

givi [52]

The solid compound, K2SO4 contains a cation called K+ and an anion called SO42-. In this case, there are 2 atoms of potassium, 1 atom of sulfur and 4 moles of oxygen. The compound also contains ionic bonds because of the composing non-metals and metal.

3 0

4 years ago

Sodium hydroxide, NaOH, reacts with sulfuric acid, H,SO,, in a neutralization reaction to

RSB [31]

Answer:

10.6 grams is approximately 0.10 moles. So you would need about 0.10 moles of sulfuric acid. That converts to about 9.80 grams.

Explanation:

hi girl i also wrote this in my test today so maube i hope it is correct.

mark me as brainliest if it helped you

7 0

3 years ago

Energy transformation of a kettle?

zubka84 [21]

Explanation:

<h3>Electrical energy into heat energy..</h3>

<h2>hope it helps.</h2><h2>stay safe healthy and happy...</h2>

6 0

3 years ago

Read 2 more answers

This table has information about the heat of fusion and the heat of vaporization of different substances. A 3-column table with

blagie [28]

H₂S

<h3>Further explanation</h3>

Given

ΔH fusion and ΔH vaporization of different substances

Required

The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes

Solution

We can use the formula :

$\tt \Delta H=\dfrac{Q}{n}$

Q=heat/energy absorbed

n = moles

The heat absorbed : 58.16 kJ

moles = 3.11

so ΔH vaporization :

$\tt \Delta H_{vap}=\dfrac{58.16~kJ}{3.11~mol}\\\\\Delta H_{vap}=18.7~kJ/mol$

The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S

(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)

3 0

3 years ago

Read 2 more answers

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

3 years ago

30 pts for all answers pls​

30 pts for all answers pls