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satela [25.4K]
1 year ago
11

The specific heat of water is 4.18 J/g•°C. How much heat does 225.0 g of water release when it cools from 85.5°C to 50.0°C? Ente

r your answer below, in whole numbers, without any commas.
Chemistry
2 answers:
Nana76 [90]1 year ago
8 0

Answer:

33400 J

Question:

The specific heat of water is 4.18 J/g•°C. How much heat does 225.0 g of water release when it cools from 85.5°C to 50.0°C? Enter your answer below, in whole numbers, without any commas.

Sergio [31]1 year ago
4 0
The answer is 33400J
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To maximize the yield in a certain manufacturing process, a solution of a weak monoprotic acid that has a concentration between
stira [4]

Answer: C 115 ml

Explanation:

If you have a solution of a monoprotic acid, it means that it has the form HA (just one Hydrigen atom). Therefore, one molecule of acid is going to react with just one molecule of  NaOH.

So, if you have a solution of 100 ml of 0,2 HA acid, it is going to react with a 100 ml of 0,2 M NaOH solution. Beacause we know that the acid can be a little more concentrated than 0,2 M (0,2-0,3), it probably needs a little more than 100 ml of NaOH to react.  So, the answer is C.

in 100 ml of 0,2M acid you can find 0,002 mol of HA

in 100 ml of 0,3 M acid you can find 0,003 mol of HA

in 100 ml of 0,2 M NaOH you can find 0,002 mol of NaOH

the answer can not be 200 because in 200 ml of 0,2 M NaOH there are 0,004 mol of NaOH, which is more than 0,003 mol.

0 0
2 years ago
A solution contains 0.0150 M Pb2+(aq) and 0.0150 M Sr2+(aq) . If you add SO2−4(aq) , what will be the concentration of Pb2+(aq)
AnnyKZ [126]

Answer:

\large \boxed{1.10 \times 10^{-3}\text{ mol/L}}

Explanation:

1. Concentration of SO₄²⁻

SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷

                   0.0150          x

K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}

2. Concentration of Pb²⁺

PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸

                        x          2.293 × 10⁻⁵

K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}

 

4 0
1 year ago
The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5
Natasha2012 [34]
For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
4 0
2 years ago
Complete the sentences to explain why the molecule ch3f is polar.
Lelu [443]
A group of environmentalists were discussing the benefits and drawbacks associated with using fossil fuels. Which argument <span>best </span>fits the conversation? 
Fossil fuels are cheaper than alternative forms of energy.Fossil fuel reserves will never be depleted.<span>Fossil fuels are easily renewed. </span><span>Fossil fuel use does not affect the environment.</span>
8 0
2 years ago
Read 2 more answers
By mistake, a quart of oil was dumped into a swimming pool that measures 25.0 m by 30.0 m. The density of the oil was 0.750 g/cm
kondor19780726 [428]

The oil slick thick = 1.256 x 10⁻⁴ cm

<h3>Further explanation</h3>

Volume is a derivative quantity derived from the length of the principal

The unit of volume can be expressed in liters or milliliters or cubic meters

The conversion is

1 cc = 1 cm3

1 dm = 1 Liter

1 L = 1.06 quart

<em>so for 1 quart = 0.943 L</em>

\tt 0.943~L=0.943\times 10^{-3}m^3

Volume of oil dumped = volume of swimming pool

\tt 0.943\times 10^{-3}~m^3=25\times 30\times h(h=thick)\\\\h=\dfrac{0.943\times 10^{-3}}{750~m^2}=1.257\times 10^{-6}~m=\boxed{\bold{1.256\times 10^{-4}~cm}}

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2 years ago
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