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3 years ago

An _____ is a substance that protects vulnerable molecules from free radicals.

1 answer:

4 0

I believe the correct answer is antioxidant. An antioxidant is a substance that protects vulnerable molecules from free radicals. Antioxidants are man-made or natural substances that may prevent or delay some types of cell damage due to free radicals.

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Identify the group of elements that does not readily combine with other elements.

7nadin3 [17]

Noble gases
also known as Inert gases

4 0

3 years ago

4. Juan tiene un cilindro de 40 libras herméticamente cerrado con gas propano y lo conecta mediante una válvula a otro cilindro

ch4aika [34]

Answer:

Ver explicacion

Explanation:

Cuando el cilindro de 40 libras está conectado al cilindro de 100 libras, generamos una presión que continúa hasta que los dos cilindros alcanzan la misma presión de gas en equilibrio. Recuerde que el cilindro de 100 libras estaba inicialmente vacío. Esto significa que su presión inicial es 0. El cilindro de 40 libras ya estaba lleno, por lo que dividimos esta cantidad en dos para tener en cuenta su distribución entre los dos cilindros.

Ahora tenemos 20 libras de gas propano presentes en cada cilindro. La implicación de esto es que, en el cilindro de 100 libras, necesitamos 80 libras adicionales para completar las 100 libras.

5 0

2 years ago

*WILL GIVE BRAINLIEST*

denis-greek [22]

Weathering breaks down and loosens the surface minerals of rock so they can be transported away by agents of erosion such as water, wind and ice. There are two types of weathering: mechanical and chemical. Mechanical weathering is the disintegration of rock into smaller and smaller fragments.

3 0

3 years ago

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]

Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

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

3 years ago

Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is starte

Katarina [22]

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) + CO₂ (g) → CH₄N₂O (s) + H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

3 0

3 years ago