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Lana71 [14]
3 years ago
11

13) Calculate and Round to the correct number of significant digits. *

Chemistry
1 answer:
Lesechka [4]3 years ago
7 0

Answer:

333(3significants points)

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The isomerization of methylisonitrile to acetonitrileCH3NC(g)→CH3CN(g)is first order in CH3NC . The rate constant for the reacti
lisabon 2012 [21]

Answer:

Option E, Half life = 2.96\times 10^3\ s

Explanation:

For a first order reaction, rate constant and half-life is related as:

            t_{1/2}=\frac{0.693}{k}

Where,

t_{1/2} = Half life

k = Rate constant

Rate constant given = 2.34\times 10^{-4}\ s^{-1}

t_{1/2}=\frac{0.693}{k}

=\frac{0.693}{2.34 \times 10^{-4}}=2.96\times 10^3\ s

So, the correct option is option E.

4 0
3 years ago
URGENT CHEMISTRY EXPERT!
vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl

(b) Moles of CuCl₂

\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} =  \text{3.340 mmol CuCl}_{2}

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

\text{Moles of Na$_{3}$PO}_{4} =  \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}

(d) Volume of Na₃PO₄

V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}

Part 2. Net ionic equation

(a) Molecular equation

\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)  

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>  

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0
3 years ago
2Al + 6HCl --&gt; 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L
dem82 [27]

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

<em>Moles H2:</em>

PV = nRT; PV/RT = n

<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

<em />

<em>Moles Al:</em>

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

<em />

<em>Mass Al -Molar mass: 26.98g/mol-:</em>

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries

7 0
3 years ago
How many moles of CaCl2 are contained in 0.448 L of a 0.85 M CaCl2 solution?
kati45 [8]

Answer:

0.3808

Explanation:

number of moles,n=Conc.XVol.

hence 0.85X0.448

3 0
3 years ago
A cathode ray tube is made of glass with a small amount of some kind of gas in it. It has metal electrodes at each end to pick u
Butoxors [25]

Answer:

One of the main uses of the cathode ray tube is in the Cathode  ray oscilloscope

Explanation:

Cathode rays are produced when a gas in an evacuated glass at very low pressure and high pressure breaks up into positive and negative ions. the negative ions move towards the anode(positive electrode) while the positive ions move towards the cathode(negative electrode), and there they knock off electrons (which are known as cathode rays) from the metal plate of the cathode.

Cathode ray tubes are mainly used in oscilloscopes, television picture tubes and in computer screens.

The cathode ray oscilloscope is used in a.c. and d.c. voltage measurements, observation of waveforms, time measurements, etc.

5 0
3 years ago
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