All categories

2 years ago

Products for combustion

Chemistry

2 answers:

Marrrta [24]2 years ago

8 0

7 drops of semen, 4 cups of menstrual blood and breastmilk

V125BC [204]2 years ago

7 0

Answer:

Carbon Dioxide

Carbon Monoxide

Sulfur Dioxide

Nitrogen Oxides

Lead

Particulate Matter

Explanation:

You might be interested in

Round 4,474 to the nearest thousands

Masja [62]

The answer would be 4,000
Because 474 is gonna go down but if it was 500 it will go up

4 0

3 years ago

How many moles does the mass quantity 127.4g of AgNO3 have?

IgorC [24]

The mole is 0.749 because molar mass of AgNO3=170g/mol and mole=mass/molar mass.
mole=127.4g/170g/mol

3 0

3 years ago

Identify this reaction <br><br> HBr+ Al(OH)3 —-> H2O + AlBr3

Vikki [24]

<h2><em><u>Answer</u></em></h2>

Neutralization reaction

5 0

4 years ago

A 12 letter word for a force that holds particles together in the atomic nucleus?

jeka94

Answer:

im not sure, are you sure?

3 0

3 years ago

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vess

Tpy6a [65]

Answer:

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

Explanation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The relation of $K_c\& K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure.=98.1

$K_c$ = Equilibrium constant in terms of concentration =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side= $n_{g,p}-n_{g.r}=1-2=-1$

$98.1=K_c(RT)^{-1}$

$98.1 =\frac{K_c}{RT}$

$K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}$

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

5 0

3 years ago

Read 2 more answers

Products for combustion​

Products for combustion