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monitta
2 years ago
6

45.8 grams of copper requires 1172 J of heat to reach a final temperature of 100.5 ℃. What was the initial temperature? Specific

heat of copper is 0.385 J/g℃. (7.02)
A. 66.5[?][?][?]
B. 122[?][?][?]
C. 34.0[?][?][?]
D. 75.2[?][?][?]

I also have no idea what the [?][?][?] is. Can someone please help me for 25 points? :'(
Chemistry
1 answer:
Jet001 [13]2 years ago
5 0

Answer: Option A, 66.5

Explanation:

So you want to use the heat capacity formula

So, q(heat)=m(mass) * cp (specific heat) * DeltaT (change in temp)

so m=45.8, cp =0.385, q=1172

you dont have the final temp for DeltaT so you have to fill in and divide

1172 = (45.8) * (0.385) * Delta T

1172 = 17.633 * DeltaT (Now divide)

1172/17.633 = Delta T

66.46 = Delta T, rounded you get 66.5 or option A

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Answer:

It sounds fine, but it may be a bit too long. It's difficult to shorten things like this, but getting more straight to the point would give it that "catchy" feel.

Explanation:

3 0
3 years ago
Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
The element silver exists in nature as two isotopes: 107Ag has a mass of 106.9051 u, and 109Ag has a mass of 108.9048 u. The ave
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Answer:

107Ag has abundance of 51.7%

109Ag has abundance of 48.3%

Explanation: Please see attachment for explanation

3 0
3 years ago
Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov
love history [14]

Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field  = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3

Mass of the Astroturf = m

Density of the Astroturf = d = 187 oz/ft^3

d=\frac{m}{V}

m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz

1 oz = 0.0283495 kg

m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg

Weight of the Astroturf = W

W = mg

=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N

The weight of the Astroturf is 179,684.31 Newtons.

8 0
3 years ago
What is the mass in grams of 17.46 mL m L of acetone?
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Answer:

17,46gram

Explanation:

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4 0
3 years ago
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