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3 years ago

Air pollution may have a greater effect on health than on atmospheric temperature.

1 answer:

8 0

<span>Yes it is true, Because we are breath in air in the form of smoke and very small practical plagued our lungs, Which are very danger for our body. Computerized tomography imaging studies on the bodies of ancient mummies have detected evidence of pneumonia, emphysema and many types of asthma problem occur in our body .</span>

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Calculate the molarity of H3PO4 when you added 57.3 g into 3,820 mL of water

motikmotik

Answer:

0.153M

Explanation:

57.3/97.994 (molar mass)=0.585 moles of H3PO4

.0585/3.820L=0.153M

5 0

3 years ago

A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and

Dmitry_Shevchenko [17]

Answer:

1. $Rate =k [NO]^{2}[Cl_{2}]$

2. $k= 0.42 \frac{L^{2}}{mol^{2}*s}$

Explanation:

$Rate =k [NO]^{m}[Cl_{2}]^{n}$

$Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2$

$Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1$

$Rate =k [NO]^{2}[Cl_{2}]^{1}$

$Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}$

6 0

3 years ago

What was a Parthian battery?

timama [110]

Answer:

Scientists believe the batteries (if that is their correct function) were used to electroplate items such as putting a layer of one metal (gold) onto the surface of another (silver), a method still practiced in Iraq today.

Explanation:

8 0

3 years ago

Please help with any

lys-0071 [83]

10 down: Particle with no charge

Answer: Neutron

3 0

3 years ago

PLEASE HELP!!!

trasher [3.6K]

Answer:

(a) oxygen

(b) 154g (to 3sf)

Explanation:

mass (g) = moles × Mr/Ar

note: eqn means chemical equation

(a)

moles of P = 84.1 ÷ 30.973 = 2.7152 moles

moles of O2 = 85÷2(16) = 2.65625 moles

Assuming all the moles of P is used up,

moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)

moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)

therefore there is insufficient moles of O2 and the limiting reactant is oxygen.

(b)

moles of P2O5 produced

= 2/5 (according to eqn) × 2.7152

= 1.08608moles

mass of P2O5 produced

= 1.08608 × [ 2(30.973) + 5(16) ]

= 154.164g

= approx. 154g to 3 sig. fig.

(c)

% yield = actual/theoretical yield × 100%

= 123/154 × 100%

= 79.870%

= approx. 79.9% (to 3sf)

4 0

3 years ago