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3 years ago

PLEASE HELP THIS IS SCIENCE

Chemistry

2 answers:

timofeeve [1]3 years ago

7 0

I'm pretty sure the answer is A.

dlinn [17]3 years ago

4 0

Answer:

where is the answer??

Please do not just take the points without awnsering the question.

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Your local grocery store

umka2103 [35]

The total number of matchbooks required would 5,000 be and the cost would be $144

<h3>Mathematical ratios</h3>

50 matchbooks = $1.44

Each matchbook = 0.005 grams red phosphorus

25 grams of red phosphorus is needed:

25/0.005 = 5,000 matchbooks

5,000 matchbooks will cost:

5000/50 x 1.44 = $144

More on mathematical ratios can be found here: brainly.com/question/20387079

#SPJ1

3 0

2 years ago

How many grams of KCl are there for in 10 moles of salt?

galben [10]

Your first step is determining how many gram of KCl are in every mole of KCl. This can be done by simply looking at K and Cl's atomic mass on the Periodic Table. You add K's atomic mass (39.1g) with Cl's atomic mass (35.45g) to determine that the mass of one mole of KCl is 74.55g. Because you have 10 moles of KCl you multiply 74.55g by 10 to reach your answer of 745.5g.

4 0

4 years ago

ZILLDIFFEQMODAP11 3.1.021. My Notes Ask Your Teacher A tank contains 150 liters of fluid in which 10 grams of salt is dissolved.

SSSSS [86.1K]

Answer:

Therefore, the number of grams of salt in the tank at time t is $A(t) = 150-140 e^{-\frac{t}{30} }$

Explanation:

Given:

Tank A contain $V_{1} = 150$ lit

Rate $\alpha = 5 \frac{L}{min}$

Dissolved salt $A = 10$ gm

Salt pumped in one minute is $4 \frac{L}{min}$

Salt pumped out is $\frac{5L}{150L} = \frac{1}{30}$ of initial amount added salt.

To find $A(t)$

$\frac{dA}{dt} = Rate _{in} - Rate _{out}$

$A' = 5 - \frac{A}{30}$

$A' + \frac{A}{30} = 5$

Solving above equation,

$I .F = e^{\int\limits {p} \, dt }$

$y = e^{\int\limits {\frac{1}{30} } \, dt }$

$y = e^{\frac{t}{30} }$

$(Ae^{\frac{t}{30} } )' = 5 e^{\frac{t}{30} } + c$

Integrating on both side,

$Ae^{\frac{t}{30} } = 5 \times 30 e^{\frac{t}{30} } +c$

Add $e^{-\frac{t}{30} }$ on above equation,

$A = 150 + ce^{-\frac{t}{30} }$

Here given in question, $A(t=0) = 10$

$10 =150 +c$

$c = -140$

Put value of constant in above equation, and find the number of grams of salt in the tank at time t.

$A(t) = 150-140 e^{-\frac{t}{30} }$

Therefore, the number of grams of salt in the tank at time t is $A(t) = 150-140 e^{-\frac{t}{30} }$

7 0

3 years ago

How many molecules are in 13.5 g of sulfur dioxide?

vampirchik [111]

Oxygen-8×2=16, sulphur=16
1mole=6.023×10^23molecule=22.4l=32gram

32g=6.023×10^23molecule
1g=6.023×10^23÷32
13.5g=6.023×10^23÷32×13.5
=2.5×10^23 molecule
.°. there are 2.5×10^23 molecule in 13.5g of sulphur dioxide

4 0

3 years ago

Read 2 more answers

An internal combustion engine relies primarily on the efficient production of energy by the combination of oxygen (02) and gasol

jeka57 [31]

Answer is: <span>the exact ratio of oxygen to octane for is 12.5 : 1.
</span>Balanced chemical reaction: C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O or multiply by 2:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O.
There same number of atoms on both side of balanced chemical reaction: eight carbon atoms, eighteen hydrogen atoms and twenty five oxygen atoms.

7 0

3 years ago