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4 years ago

What temperature scale has no negative numbers?

Chemistry

1 answer:

Mamont248 [21]4 years ago

8 0

Kelvin

When measuring in kelvin units the lowest unit is 0k

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1. A cupcake show makes and delivers specialized cupcakes. It will cost someone $30 for

lorasvet [3.4K]

Answer:

2.5

Explanation:

The given costs for the purchase and delivery of the cupcakes are;

The cost of 10 cupcakes and the delivery fee = $30

The cost of 25 cupcakes and the delivery fee = $67.50

Let 'c' represent the fixed cost of delivery of the cup cakes, and let 'x' represent the cost of each cup cake, we have;

30 = 10·x + c...(1)

67.50 = 25·x + c...(2)

Subtracting equation (1) from equation (2) gives;

67.50 - 30 = 37.50 = 25·x - 10·x + c - c = 15·x

∴ x = 37.50/15 = 2.5

The cost of each cupcake, x = 2.5

(∴ The delivery fee = 30 - 10 × 2.5 = 5)

The cost of each cupcake is 2.5.

8 0

3 years ago

What is the empirical formula for C2H6?<br> CH<br> CH3<br> C2H4<br> C6H18

Ket [755]

Answer: CH3

Explanation: Because if we double each element count, we arrive at C2H6. CH3 is the empirical formula.

5 0

3 years ago

The average us farm occupies 435acres.how many square miles is this

svetlana [45]

435 acres is about 0.679688 hope this helped!

8 0

3 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

to save time you can approximate the initial volume of water to ±1 ml and the initial mass of the solid to ±1 g . for example, i

makkiz [27]

The correct answer for this question is

20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.

15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.

11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.

To know more about Density, click here:

brainly.com/question/6107689

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8 0

2 years ago