A store sells 4 reams of paper for $23.08. Which proportion can be used to find the cost of 12 reams of paper?

Sue is building a bike with a metal frame in a warm, sunny area. Which material would be best for Sue to use for the grips on th

Ede4ka [16]

Answer:

Rubber, or another insulator.

Explanation:

I'm not sure what the options are, but Sue would probably want an insulator so that the heat is trapped, keeping her hands cool.

6 0

3 years ago

Read 2 more answers

Read the following chemical equations.

lora16 [44]

This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:

Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰

Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻

Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.

In such a way, the correct choice is C.

Learn more:

(Redox reactions) brainly.com/question/13978139

8 0

2 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

4 years ago

2. Answer this question: Lava rocks I nature come in different shapes and sizes. What processes do you think are responsible the

GuDViN [60]

Answer: im not sure im only in 8th grade but im pretty sure Erosion i started learning about this last year i really cant explain it... im still hainving trouble with it.

Explanation:

4 0

3 years ago

Mrs. Stark weighs 70kg. How much Potential Energy does she have if she stands on the roof of an 80m tall building?

8_murik_8 [283]

E = mgh

E = 70*9.81*80

E = 54936 J

5 0

3 years ago