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ad-work [718]
4 years ago
12

A projectile moves horizontally with ____?____motion HELP PLEASE

Chemistry
1 answer:
blagie [28]4 years ago
3 0

Answer:

hi

Explanation:

hi

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Meanders, water<span> flows fastest on the outer bend of the river where the channel is deeper and there is less friction.</span>
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3 years ago
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Blast furnaces extra pure iron from the Iron(IIl)oxide in iron ore in a two step sequence. In the first step, carbon and oxygen
OLga [1]

Answer:

5.9 kg  

Explanation:

We must work backwards from the second step to work out the mass of oxygen.

1. Second step

Mᵣ:                                     55.84

            Fe₂O₃ + 3CO  ⟶  2Fe  +  3CO₂

m/kg:                                    7.0

(a) Moles of Fe

\text{Moles of FeO} = \text{7000 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.84 g Fe}} = \text{125 mol Fe}

(b) Moles of CO

\text{Moles of CO} = \text{125 mol Fe} \times \dfrac{\text{3 mol CO}}{\text{2 mol Fe}} = \text{188 mol CO}

However, this is the theoretical yield.

The actual yield is 72. %.

We need more CO and Fe₂O₃ to get the theoretical yield of Fe.

(c) Percent yield

\begin{array}{rcl}\text{Percent yield} &=& \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \, \%\\\\ 72. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.72 &= &\dfrac{\text{188 mol}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{188 mol}}{0.72}\\& = & \textbf{261 mol}\\\\\end{array}

We must use 261 mol of CO to get 7.0 kg of Fe.

2. First step

Mᵣ:                32.00

            2C   +  O₂   ⟶  2CO

n/mol:                             261

(a) Moles of O₂

\text{Moles of O}_{2} = \text{261 mol CO} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol CO}} = \text{131 mol O}_{2}

(b) Mass of O₂

\text{Mass of O}_{2}= \text{131 mol O }_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol  O}_{2}} = \text{4180 g O}_{2}

However, this is the theoretical yield.

The actual yield is 71. %.

We need more C and O₂ to get the theoretical yield of CO.

(c) Percent yield

\begin{array}{rcl}71. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.71 &= &\dfrac{\text{4180 g}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{4180 g}}{0.71}\\\\& = & \text{5900 g}\\& = & \textbf{5.9 kg}\\\end{array}

We need 5.9 kg of O₂ to produce 7.0 kg of Fe.

6 0
4 years ago
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3 0
3 years ago
When 0.040 mol of propionic acid, c2h5co2h, is dissolved in 750 ml of water, the equilibrium concentration of h3o+ ions is measu
torisob [31]

<span>Answer is: Ka for propinoic acid is 6,57·10</span>⁻⁵.<span>
Chemical reaction: C</span>₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).<span>
n(C</span>₂H₅COOH) = 0,04 mol.<span>
V(C</span>₂H₅COOH) = 750 mL = 0,75 L.<span>
c(C</span>₂H₅COOH) = 0,04 mol ÷ 0,75 L.<span>
c(C</span>₂H₅COOH) = 0,053 mol/L = 0,053 M.<span>
[C</span>₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [C</span>₂H₅COO⁻] · [H₃O⁺] / [C₂H₅COOH].<span>
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10</span>⁻⁵.

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3 years ago
make a chart to show the effect of Sodium Hydroxide and Sulphuric acid on methyl orange and red litmus paper​
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Answer:

mark me as brainliest plsss

3 0
3 years ago
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