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2 years ago

Which of the graphs below shows a person walking at a constant rate with no change in speed or velocity?

Chemistry

2 answers:

charle [14.2K]2 years ago

8 0

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The graph showing a person who is walking at the same rate with no change in speed or velocity is :

Person B

maria [59]2 years ago

7 0

Answer:

all of these maybe I think

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The lab procedure mentions that not all of the oxone may dissolve after being added to the water because the water may be satura

ollegr [7]

Answer:

The answer is (C) There are more solute molecules than water molecules.

Explanation:

A saturated solution is one in which no more solute can be dissolved or disintegrated into the solvent. When or if the ozone stops being dissolved in the water, it implies that the water has already taken on more ozone molecules than it can contain, meaning there are more solute molecules (ozone molecules) than there are solvent molecules (water molecules).

3 0

2 years ago

Read 2 more answers

Which of the following is not a correct chemical equation for a double displacement reaction?

Step2247 [10]

Answer:

B. CaCl + LiCO3 yields CaCO3 + LiCl is not correct

It should be CaCl2 + Li2CO3 → 2LiCl + CaCO3

Explanation:

For a reaction to be double displacement reaction there are two things we need to look for

1) There must be an interchange of the group of ions

2) The reactants must dissolve in water to release ions

A. 2RbNO3 + BeF2 yields Be(NO3)2 + 2RbF

2Rb+ + NO3- + Be^2+ + 2F- → Be(NO₃)₂ + 2RbF

This is correct

B. CaCl + LiCO3 yields CaCO3 + LiCl

This is not correct

The correct equation is:

CaCl2 + Li2CO3 → Ca2+ + 2Cl- + 2Li+ + CO3^2- → 2LiCl + CaCO3

C. Na3PO4 + 3KOH yields 3NaOH + K3PO4

3Na+ + PO4^3- + 3K+ + 3OH- → 3NaOH + K3PO4

This is correct

D. 2MgI2 + Mn(SO3)2 yields 2MgSO3 + MnI4

2Mg^2+ + 4I- + Mn^4+ + 2SO3^2- → 2 MgSO3 + MnI4

This is correct

8 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

2 years ago

a sample of ammonia liberates 5.66 kj of heat as it solidifies at its melting point . what is the mass of the sample?

Reil [10]

The correct answer for the question that is being presented above is this one: "16.728 g."

Given that
ΔHsolid = -5.66 kJ/mol.
This means that 5.66 kJ of heat is released when 1 mole of NH3 solidifies

When 5.57 kJ of heat is released
amount of NH3 solidifies = 5.57/5.66 = 0.984 moles

molar mass of NH3 = 17 g/mole 
1 mole of NH3 = 17 g 
So, 0.984 moles of NH3 = 17 X 0.984 = 16.728 g

7 0

3 years ago

MgNH4PO4•6H2O loses H2O in a stepwise fashion as it is heated. Between 40 degrees celsius and 60 degrees celsius, the monohydrat

tino4ka555 [31]

Answer:

$\%\ Composition\ of\ phosphorus=20.00\ \%$