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Savatey [412]
3 years ago
5

Seven vans each hold 6 passengers. Each passenger is carrying 20 candies divided equally among 5 small boxes. How many candies a

re on the vans? 840 candies 4200 candies 168 candies 120 candies
Chemistry
1 answer:
HACTEHA [7]3 years ago
8 0
I believe the correct answer from the choices listed above is the first option. There are about 840 candies present on all vans present. We calculate it by multiplying the number of total passengers by the number of candies each passenger is carrying. Hope this answers the question.
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(Please help, ASAP) How many grams are in 3.45x10^23 atoms of P?
Daniel [21]

one mole of P weights about 31 grams

in one mole there are 6.022*10^23 atoms

we use the rule of threes

6.022*10^23atoms......weight..........31 grams

3.45*10^23 atoms.........weight...........x grams

x=(3.45*10^23*31)/6.022*10^23

x=106.95/6.022=<u><em>17.76 grams</em></u>

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3 years ago
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Answer:

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Explanation:

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34. 3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases
Digiron [165]

Answer:

ΔH = 2.68kJ/mol

Explanation:

The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:

q = m*S*ΔT

<em>Where q is heat of reaction in J,</em>

<em>m is the mass of the solution in g,</em>

<em>S is specific heat of the solution = 4.184J/g°C</em>

<em>ΔT is change in temperature = 11.21°C</em>

The mass of the solution is obtained from the volume and the density as follows:

150.0mL * (1.20g/mL) = 180.0g

Replacing:

q = 180.0g*4.184J/g°C*11.21°C

q = 8442J

q = 8.44kJ when 3.15 moles of the solid react.

The ΔH of the reaction is:

8.44kJ/3.15 mol

= 2.68kJ/mol

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3 years ago
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Ivanshal [37]

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3 years ago
Read 2 more answers
if 0.3 moles of carbon reacts completely with 0.3 moles of molecular oxygen, what is the yield in Percent of 6.6 g of CO2 are fo
Anettt [7]
Answer is:<span>the yield is 50%.
</span>
Chemical reaction: C + O₂ → CO₂.
n(C) = 0.3 mol; amount of substance.
n(O₂) = 0.3 mol.
From chemical reaction: n(C) : n(CO₂) = 1 : 1.
n(CO₂) = 0.3 mol.
M(CO₂) = 44 g/mol; molar mass of caron(IV) oxide.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) =0.3 mol · 44 g/mol.
m(CO₂) = 13.2 g; mass of carbon(IV) oxide.
the yield = 6.6 g ÷ 13.2 g · 100%.
the yield = 50%.
8 0
3 years ago
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