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3 years ago

7

Each Thursday the 11 kindergarten students in Miss Goodson's class are each allowed one slice of pie, one cup of orange juice, a

nd two doughnut holes. The leftovers will be given to the custodian on the night shift.
How many slices of pie are left for the custodian?
slices of pie
How many cups of orange juice are left for the custodian?
cups of orange juice
How many doughnut holes are left for the custodian?

donut holes

1 answer:

mixas84 [53]3 years ago

4 0

How many were there to start off with?

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A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

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3 years ago

What volume of Co2 (carbon (iv) oxide)

hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

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4 years ago

Anh measured the temperature of a pond near his house. Before he left for school, the water in the pond was 18 degrees celsius.

Hitman42 [59]

As the temperature increases, B) the molecules started moving faster.

Explanation:

The temperature of a substance is a measure of the average kinetic energy of the particles in a substance. In particular, it can be found that the temperature is directly proportional to the average kinetic energy of the particles:

$T\propto KE$

The kinetic energy of a particle is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its speed

This means that the higher the temperature of a substance, the greater the speed of the particles in the substance.

Therefore, if we apply this concept to this problem, we infer that as the temperature of the water in the pond gets higher, the speed of the molecules inside the water increases, which means that the molecules are moving faster.

Therefore, the correct answer is

B) the molecules started moving faster.

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4 years ago

copper forms two oxides. On heating 1 g of each in hydrogen 0.888 g and 0.798 g of the metal was obtained. Show that the results

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Answer:

For the first oxide, 1 g gives 0.888 g of copper.

Dividing by 0.888 tells us that 1.126 g gives 1 g of copper so has 0.126 g of oxygen.

For the second oxide, 1 g gives 0.798 g of copper.

Dividing by 0.798 tells us that 1.253 g gives 1 g of copper so has 0.253 g of oxygen.

So 1 g of copper combines with either 0.126 g or 0.253 g of oxygen.

Within the limits of experimental error, 0.253 is twice 0.126, confirming the law of multiple proportion.

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