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melomori [17]
3 years ago
15

When determining the density of a grape, a student was not careful when putting the grape into the graduated cylinder, and some

of the water got splashed out. How would that affect the determined density of the grape? In other words, is the calculated density be higher or lower than the actual density of the grape? Explain your answer (fill in the blanks below).
If some of the water got splashed out, the measured “grape + water” volume is incorrectly ______________________ (high or low?).
Then the calculated volume of the grape (“grape+water” – “water only” volume) is incorrectly _______________ (high or low?).
When calculating the density, the mass is divided by a (higher or lower?) ______________ value. As the result, the determined density is erroneously (high or low?) _____________ and therefore the determined density is (higher or lower?) ______________than the actual density of the grape.
Chemistry
1 answer:
galina1969 [7]3 years ago
4 0

Answer:

gtjgyjtjyjjjjk

Explanation:

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Calculate the inner diameter of the steam tube between the boiler and the high pressure turbine if the average velocity of the s
stich3 [128]

Answer:

The inner diameter of the steam tube is 0.1185 m.

Explanation:

m=\rho \times A\times v=\frac{1}{V}\times A\times v

m = mass flow rate

\rho = density of the liquid

V=specific volume =\frac{1}{\rho }

v = velocity of the liquid

A = area of cross-section

Diameter of the cross-section = d = 2r

Given: m = 18 kg/s, v = 60 m/s,A=?, V = 0.036793 m^3/kg

18 kg/s=\frac{1}{0.036793 m^3/kg}\times A\times 60 m/s

A=0.0110379 m^2

Area of the inner cross section = \pi r^2=\pi \frac{d^2}{4}

\pi \frac{d^2}{4}=0.0110379 m^2

d = 0.1185 m

The inner diameter of the steam tube is 0.1185 m.

5 0
4 years ago
Imagine a made-up pol
zheka24 [161]

Answer:

No change to the cation  Add -ide to the anion

Question 1 and .2

Explanation:

6 0
3 years ago
6. There are three isotopes of silicon. They have mass numbers of 28, 29 and 30. The average atomic mass of silicon is 28.086amu
Elza [17]

Answer:

That means that the isotope that has a mass number of 28 is probably the most abundant. This is because your average atomic mass is 28.086 amu, which is closest to 28.

Formula used to calculate average atomic mass follows:

average atomic mass=(atomic mass of an isotope)*(fractional abundance)

we are given=

Three isotopes of Silicon, which are Si-28, Si-29 and Si-30.

Average atomic mass of silicon = 28.086 amu

As, the average atomic mass of silicon is closer to the mass of Si-28 isotope. This means that the relative abundance of this isotope is the highest as compared to the other two isotopes.

Percentage abundance of Si-28 isotope = 92.2%

Percentage abundance of Si-29 isotope = 4.7 %

Percentage abundance of Si-30 isotope = 3.1%

Hence, the abundance of Si-28 isotope is the highest as compared to the other isotopes.

7 0
3 years ago
Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti
ss7ja [257]

Answer:

1.89 of Sodium Carbonate

3.94 g of Silver Carbonate

2.43 g of Sodium Nitrate

Zero grams of Silver Nitrate

Explanation:

We have to start with the reaction:

AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~NaNO_3

Now, we can balance the reaction:

2AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~2NaNO_3

Now, we have to calculate the limiting reagent and we have to follow a few steps:

1) Convert to moles (using the molar mass of each compound)

2) Divide by the coefficient of each reactive (given by the balanced reaction)

<u>Convert to moles</u>

<u />

3.40~g~Na_2CO_3\frac{105.98~g~Na_2CO_3}{1~mol~Na_2CO_3}=0.032~mol~Na_2CO_3

4.86~g~AgNO_3\frac{169.8~g~AgNO_3}{1~mol~AgNO_3}=0.0286~mol~AgNO_3

<u>Divide by the coefficient</u>

<u></u>

\frac{0.032~mol~Na_2CO_3}{1}=0.032

<u />

\frac{0.0286~mol~AgNO_3}{2}=0.0143

The smallest value is for AgNO_3 , therefore the 4.86 g of AgNO_3 .

Now we can calculate the amount of compounds produced is we follow a few steps:

1) Use the molar ratio

2) Convert to moles (using the molar mass of each compound)

<u>Amount of Silver Carbonate</u>

<u />

0.0286~mol~AgNO_3\frac{1~mol~AgCO_3}{2~mol~AgNO_3}\frac{275.74~g~AgCO_3}{1~mol~AgCO_3}=3.94~g~AgCO_3

<u>Amount of Sodium Nitrate</u>

<u />

0.0286~mol~AgNO_3\frac{2~mol~NaNO_3}{2~mol~AgNO_3}\frac{84.99~g~NaNO_3}{1~mol~NaNO_3}=2.43~g~NaNO_3

<u>Amount of Sodium Carbonate (Excess reactive)</u>

<u />

0.0286~mol~AgNO_3\frac{1~mol~NaCO_3}{2~mol~AgNO_3}\frac{105.98~g~NaCO_3}{1~mol~NaCO_3}=1.51~g~NaCO_3

3.4~g~NaCO_3-1.51~g~NaCO_3=1.89~g~NaCO_3

<u>Amount of Silver Nitrate</u>

<u />

All the silver nitrate would be consumed in the reaction

I hope it helps!

<u />

<u />

<u />

7 0
4 years ago
Please help I will give brainliest to whoever answers it correctly
UkoKoshka [18]

Answer:

5.477 rounded to 5.5

Explanation:

4 0
4 years ago
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