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3 years ago

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A 50.0-g sample of liquid water at 25.0 Â°c is mixed with 23.0 g of water at 79.0 Â°c. the final temperature of the water is ___

_____ Â°c.

1 answer:

I am Lyosha [343]3 years ago

8 0

<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>

<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>

<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>

<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>

<span>73.0<span>Tf</span>=2561 °C</span>

<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>

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A gas at STP occupies 22.4 L if the temperature is changed to 260 K and the pressures changed it to 0.50 ATM what will the new v

asambeis [7]

Answer:

The new volume will be 42, 7 L.

Explanation:

We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. The conditions STP are: 1 atm of pressure and 273 K of temperature.

P1xV1/T1 =P2xV2/T2

1 atmx 22,4 L/273K = 0,5atmx V2/260K

V2=((1 atmx 22,4 L/273K )x 260K)/0,5 atm= 42, 67L

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3 years ago

What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.

AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

$pH=pK_{a} + log(\frac{[Base]}{[Acid]})$

According to the given conditions, the equation will become as follow

$pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})$

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

$pH=6.86 + log(\frac{0.058}{0.042})$

$pH=7.00$

5 0

3 years ago

The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha

zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$25.9 = x*10.0$

$25.9 = 10.0\:x$

$10.0\:x = 25.9$

$x = \dfrac{25.9}{10.0}$

$\boxed{x = 2.59}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{53.3}{2^{2.59}}$

$m \approx \dfrac{53.3}{6.021}$

$\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

3 0

3 years ago

The intermolecular forces between polymer molecules are strong, so these substances are __________ at room temperature. What wor

geniusboy [140]

Answer:

I want to say

Stronger

8 0

2 years ago

Lithium bromide dissociates in water according to the following thermochemical equation: LiBr(s) → Li+ (aq) + Br– (aq) ΔH = –48.

madreJ [45]

Answer:

Final temperature of water is $48.3^{0}\textrm{C}$

Explanation:

1 mol of LiBr releases 48.83 kJ of heat upon dissolution in water.

So, 2 moles of LiBr release $(2\times 48.83)kJ$ or 97.66 kJ of heat upon dissolution in water.

This amount of heat is consumed by 1000.0 g of water. Hence temperature of water will increase.

Let's say final temperature of water is $t^{0}\textrm{C}$ .

So, change in temperature ( $\Delta T$ ) of water is $(t-25)^{0}\textrm{C}$ or (t-25) K

Heat capacity (C) of water is $4.184\frac{J}{g.K}$

Hence, $m_{water}\times C_{water}\times \Delta T_{water}=97.66\times 10^{3}J$

where m is mass

So, $(1000.0g)\times (4.184\frac{J}{g.K})\times (t-25)K=97.66\times 10^{3}J$

or, $t=48.3$

Hence final temperature of water is $48.3^{0}\textrm{C}$

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3 years ago