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2 years ago

Convert 24.3 cm to mi show equation

Chemistry

1 answer:

Lostsunrise [7]2 years ago

8 0

Answer:

if you are talking about miles the answer would be is 0.0001509932(divided by 160934)

if you are talking about mililiters it would be 24.3(multiply times one)

hope this helps

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A compound contains nitrogen and a metal. This compound goes through a combustion reaction such that compound X is produced from

tia_tia [17]

Answer:

The correct answer is: X is nitrogen dioxide, and Y is a metal oxide

Explanation:

Combustion of compound of containing nitrogen and metal will give nitrogen dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.

The general equation is given as:

$4M_3N_x+7xO_2\rightarrow 4xNO_2+6M_2O_x$

Hence, the correct answer is :X is nitrogen dioxide, and Y is a metal oxide.

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3 years ago

How many kilocalories are needed to vaporize 5.8 mol of Br2

Valentin [98]

<span>The vaporization of br2 from liquid to gas state requires 7.4 k/cal /mol.</span>

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2 years ago

Why is the first jar of hydrogen gas collected during preparation discarded

valentina_108 [34]

Answer:

Hydrogen is collected by downward displacement of water as it is less denser than water. Hence it comes out at the surface of water . It is insoluble in water so it does not dissolve in water

7 0

2 years ago

Read 2 more answers

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

2 years ago

A client who weighs 70 kg is receiving a solution of 0.9% sodium chloride (normal saline) 500 ml with dopamine 800 mg at 5 ml/ho

nordsb [41]

The mcg/kg/minute the client is receiving is 1.904 mcg/kg/min whose weight is 70 kg.

1 milligram (mg) = 1000 micrograms (mcg) or 0.001 grams (g) 1 g = 1000 mg 1 kilogram (kg) = 1000 g 1 kg = 2.2 pound (lb) 1 liter (L) = 1000 milliliters (mL)

The number of mcg/kg/minute = flow rate × concentration ÷ mass of client

Flow rate = 12 ml/hour = 12 ml/hour × 1 hr/60 min = 0.2 ml/min

Concentration = mass of dopamine/volume

where

mass of dopamine = 800 mg and

volume = 500 ml

Concentration =800 mg/500 ml

= 1.6 mg/ml

= 1.6 mg/ml × 1000 mcg/mg

= 1600 mcg/ml

mass of client = 70 kg

Calculating mcg/kg/minute

So, substituting the variables into the equation, we have

mcg/kg/minute = flow rate × concentration ÷ mass of client

mcg/kg/minute = 0.08 ml/min × 1600 mcg/ml ÷ 70 kg

mcg/kg/minute = 320 mcg/min ÷ 70 kg

mcg/kg/minute = 1.904 mcg/kg/min

Thus, the mcg/kg/minute the client is receiving is 1.904 mcg/kg/min

Learn more about mcg/kg/minute here:

brainly.com/question/4253005

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1 year ago