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2 years ago

9

Vegetarians do not eat animal products but rather derive their protein from plant sources. Which type of protein do vegetarian f

ood sources lack?
_________

1 answer:

Whitepunk [10]2 years ago

6 0

Answer: lysine and methionine

Explanation:

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What are the two ways light moves through space? for astronomy the solar system

Mariana [72]

Answer:

The Solar System moves through the galaxy with about a 60° angle between the galactic plane and the planetary orbital plane. The Sun appears to move up-and-down and in-and-out with respect to the rest of the galaxy as it revolves around the Milky Way

Explanation:

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7 0

2 years ago

Read 2 more answers

How many moles of N are in .235g of N2O

makkiz [27]

Given the molar mass of Nitrogen is 14.01g/mol you can use that to solve for the moles of nitrogen.
0.235g(1mol/14.01g) = .0168 moles.

6 0

3 years ago

You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution

vfiekz [6]

Answer:

450. g of 0.173 % KCN solution contains 779 mg of KCN.

Explanation:

Mass of the solution = m

Mass of the KCN in solution = 779 mg

Mass by mass percentage of KCN solution = 0.173%

$(m/m)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$0.173\%=\frac{779 mg}{m}\times 100$

$m=\frac{779 mg}{0.173}\times 100= 450,289 mg$

1 mg = 0.001 g

m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g

450. g of 0.173 % KCN solution contains 779 mg of KCN.

6 0

3 years ago

Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ

JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

$\frac{1}{2}$ Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

average bond enthalpy is $\frac{1134.9}{4}$ = 283.725 kJ ⋅ mol − 1

4 0

3 years ago

For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

SVETLANKA909090 [29]

Answer:

$\frac{1}{24}$

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = $\frac{1}{6}$

So, fraction of the whole athletic field reserved for each fifth class = $\frac{1}{4}(\frac{1}{6})=\frac{1}{24}$

3 0

3 years ago