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Which ONE of the following is the correct relationship between vectors P, Q and R?

Physics

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Part 1.4· 2 steps

  1. Analyze the vector diagram.

    The diagram shows vector P starting at the same point as vector Q, and ending at the same point as vector R. The head of vector Q meets the tail of vector R. This arrangement represents the head-to-tail method of vector addition.

  2. Apply the head-to-tail rule for vector addition.

    According to the head-to-tail rule, if vectors are arranged such that the head of one vector connects to the tail of the next, their sum is the vector drawn from the tail of the first vector to the head of the last vector. In this case, the resultant vector P connects the tail of Q to the head of R.
    Therefore, P⃗=Q⃗+R⃗\vec{P} = \vec{Q} + \vec{R}.

    The correct relationship is A P=Q+R\boxed{\text{A } P = Q + R}.

Part 1.5· 3 steps

  1. Determine the moles of ammonia (NH3\text{NH}_3) produced.

    At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 dm322.4 \text{ dm}^3.
    Given volume of NH3=22.4 dm3\text{NH}_3 = 22.4 \text{ dm}^3.

    Moles of NH3=Volume of NH3Molar volume at STP=22.4 dm322.4 dm3/mol=1 mol\text{Moles of NH}_3 = \frac{\text{Volume of NH}_3}{\text{Molar volume at STP}} = \frac{22.4 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} = 1 \text{ mol}

  2. Use the mole ratio from the balanced chemical equation to find the moles of hydrogen (H2\text{H}_2) required.

    The balanced equation is N2(g)+3H2(g)→2NH3(g)\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}).
    From the equation, 3 moles of H2\text{H}_2 react to produce 2 moles of NH3\text{NH}_3.

    Moles of H2=Moles of NH3×3 mol H22 mol NH3=1 mol NH3×32=1.5 mol H2\text{Moles of H}_2 = \text{Moles of NH}_3 \times \frac{3 \text{ mol H}_2}{2 \text{ mol NH}_3} = 1 \text{ mol NH}_3 \times \frac{3}{2} = 1.5 \text{ mol H}_2

  3. Calculate the volume of hydrogen (H2\text{H}_2) required at STP.

    Volume of H2=Moles of H2×Molar volume at STP=1.5 mol×22.4 dm3/mol=1.5×22.4 dm3\text{Volume of H}_2 = \text{Moles of H}_2 \times \text{Molar volume at STP} = 1.5 \text{ mol} \times 22.4 \text{ dm}^3/\text{mol} = 1.5 \times 22.4 \text{ dm}^3

    The correct option is C 1.5×22.4 dm3\boxed{\text{C } 1.5 \times 22.4 \text{ dm}^3}.

Part 1.6

We need to find which option contains 6.02×10236.02 \times 10^{23} ATOMS. This is Avogadro's number (NAN_A) of atoms, which means 1 mole of atoms.

  • A: 1 mole of CO2\text{CO}_2 1 mole of CO2\text{CO}_2 contains 1 mole of C atoms and 2 moles of O atoms. Total moles of atoms = 1+2=31 + 2 = 3 moles of atoms. Number of atoms = 3×NA3 \times N_A.

  • B: 16 g of O2\text{O}_2 gas Molar mass of O2=2×16.00=32.00 g/mol\text{O}_2 = 2 \times 16.00 = 32.00 \text{ g/mol}. Moles of O2=16 g32 g/mol=0.5 mol O2\text{O}_2 = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ mol } \text{O}_2. Since 1 mole of O2\text{O}_2 contains 2 moles of O atoms, 0.5 moles of O2\text{O}_2 contains 0.5×2=10.5 \times 2 = 1 mole of O atoms. Number of atoms = 1×NA=6.02×10231 \times N_A = 6.02 \times 10^{23} atoms.

  • C: 22.4 dm322.4 \text{ dm}^3 of N2\text{N}_2 gas at STP At STP, 22.4 dm322.4 \text{ dm}^3 of any gas is 1 mole of that gas. So, this is 1 mole of N2\text{N}_2. 1 mole of N2\text{N}_2 contains 2 moles of N atoms. Number of atoms = 2×NA2 \times N_A.

  • D: 18 g of H2O\text{H}_2\text{O} Molar mass of H2O=(2×1.01)+16.00=18.02 g/mol\text{H}_2\text{O} = (2 \times 1.01) + 16.00 = 18.02 \text{ g/mol}. Moles of H2O=18 g18.02 g/mol≈1 mol H2O\text{H}_2\text{O} = \frac{18 \text{ g}}{18.02 \text{ g/mol}} \approx 1 \text{ mol } \text{H}_2\text{O}. 1 mole of H2O\text{H}_2\text{O} contains 2 moles of H atoms and 1 mole of O atoms. Total moles of atoms = 2+1=32 + 1 = 3 moles of atoms. Number of atoms = 3×NA3 \times N_A.

Option B contains 6.02×10236.02 \times 10^{23} atoms.

The correct option is B 16 g of O2 gas\boxed{\text{B } 16 \text{ g of O}_2 \text{ gas}}.

Part 1.7· 3 steps

  1. Calculate the total molar mass of water (H2O\text{H}_2\text{O}) in the compound.

    The compound is Na2SO4⋅10H2O\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}.
    Molar mass of H2O=(2×1)+16=18 g/mol\text{H}_2\text{O} = (2 \times 1) + 16 = 18 \text{ g/mol}.
    Total mass of 10H2O=10×18=180 g/mol10\text{H}_2\text{O} = 10 \times 18 = 180 \text{ g/mol}.

  2. Calculate the molar mass of the entire compound Na2SO4⋅10H2O\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}.

    Molar mass of Na2SO4\text{Na}_2\text{SO}_4:
    Na:2×23=46 g/mol\text{Na}: 2 \times 23 = 46 \text{ g/mol}
    S:1×32=32 g/mol\text{S}: 1 \times 32 = 32 \text{ g/mol}
    O:4×16=64 g/mol\text{O}: 4 \times 16 = 64 \text{ g/mol}
    Molar mass of Na2SO4=46+32+64=142 g/mol\text{Na}_2\text{SO}_4 = 46 + 32 + 64 = 142 \text{ g/mol}.
    Molar mass of Na2SO4⋅10H2O=Molar mass of Na2SO4+Molar mass of 10H2O=142+180=322 g/mol\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} = \text{Molar mass of } \text{Na}_2\text{SO}_4 + \text{Molar mass of } 10\text{H}_2\text{O} = 142 + 180 = 322 \text{ g/mol}.

  3. Calculate the percentage composition of water.

    Percentage of H2O=Mass of 10H2OMolar mass of Na2SO4⋅10H2O×100%=180322×100%\text{Percentage of H}_2\text{O} = \frac{\text{Mass of } 10\text{H}_2\text{O}}{\text{Molar mass of } \text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}} \times 100\% = \frac{180}{322} \times 100\%

    The correct expression is C 180322×100\boxed{\text{C } \frac{180}{322} \times 100}.

Part 1.8

Avogadro's Law states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. A direct consequence of this law is that one mole of any gas occupies the same volume under the same conditions of temperature and pressure.

The visible part of option A, "A mole of any gas occupies the same volume at the same...", is the beginning of the correct statement of Avogadro's Law. Assuming it completes to "temperature and pressure", it is the correct statement.

The correct statement is A A mole of any gas occupies the same volume at the same...\boxed{\text{A A mole of any gas occupies the same volume at the same...}}.

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