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navik [9.2K]
3 years ago
15

4. A serving of cereal contains about of protein 9 mg 9 g 90 g 9 kg

Chemistry
1 answer:
aleksandrvk [35]3 years ago
7 0

Answer:

probably 90 g

Explanation:

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What mass of fes is formed if 9.42 g of fe reacts with 68.0 g of s8? 8fe(s) + s8(s) → 8fes(s)?
allochka39001 [22]
From the equation, we see that the molar ratio of Fe : S required is:

8 : 1

The moles of Fe present are: 9.42/56 = 0.168
Moles of S = 68/(32 * 8) = 0.265

The molar ratio is:

1 : 1.6

Therefore, iron is the limiting reactant as it is present in a ratio lower than that required. The ratio of

Fe : FeS  is
1 : 1

So 0.168 moles of FeS will form. The mass of FeS will be:

Mass = 0.168 * (56 + 32)
Mass = 14.78 grams

14.78 grams of FeS will be formed.
4 0
3 years ago
If an object is suspended in a fluid, neither sinking nor floating, then:
marusya05 [52]
The right answer is d
6 0
3 years ago
to find the mass of 3.2 moles of calcium for carbonate which of the following is used to calculate the amount
natka813 [3]
I believe it to be g/mol of Calcium carbonate
because to finde mass...u must have grams(g) as units....
it is the only one that have g as units 

as for the first answer the avogadros number gives u the number of atoms in  one mole of calcium carbonate....
the second one is based on ... At s.t.p one mole of gas occupies 22.4 dm³⇒to find volume 
6 0
3 years ago
What is the colour of litmus in a solution of ammonium hydroxide?​
sammy [17]

Answer:

Red litmus turns blue

Explanation:

Hope it helps

7 0
3 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
3 years ago
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