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3 years ago

How many ml are in 1.25 l of a intravenous solution of 0.9% sodium chloride?

Chemistry

1 answer:

sweet-ann [11.9K]3 years ago

8 0

Mass % of NaCl = 0.9 % (m/v)

Density of 0.9 % NaCl is about 1.0 g/mL

Volume of NaCl in mL :

$1.25 L solution * \frac{1000 mL}{1 L} * \frac{0.9 g NaCl}{100 mL solution}*\frac{1 mL}{1.0 g NaCl}$

= 11.25 mL NaCl

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How does an atom-88 become a strontium ion with a +2 charge?

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Strontium atom loses 2 electrons to become an ion with 2 electrons lesser than its atom. Your answer is C.

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3 years ago

HELP WITH CHEMISTRY PLEASE!

maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

In our problem:

P1 = ??? (is needed to be calculated) and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.

2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

In our problem:

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? (is needed to be calculated)

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.

3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

In our problem:

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? (is needed to be calculated) and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.

4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; 101.325 kPa = 1.0 atm, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.

5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

In our problem:

P1 = 2200.0 mmHg and T1 = ??? (is needed to be calculated) .

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

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3 years ago

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Explain what would happen if an electron and a proton were brought near each other and then released. why would this happen?

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They would repel one another because they they have opposing charges

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Calculate the work done when an ideal gas expands isothermally and reversibly in a piston and cylinder assembly for expansion of

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Answer:

$W=5743.1077\ J$

Explanation:

The expression for the work done is:

$W=RT \ln \left( \dfrac{P_1}{P_2} \right)$

Where,

W is the amount of work done by the gas

R is Gas constant having value = 8.314 J / K mol

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P₁ is the initial pressure

P₂ is the final pressure

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P₁ = 10 bar

P₂ = 1 bar

Applying in the equation as:

$W=8.314\times 300 \ln \left( \dfrac{10}{1} \right)$

$W=300\times \:8.314\ln \left(10\right)$

$W=2.30258\times \:2494.2$

$W=5743.1077\ J$

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3 years ago

In terms of valence electrons, explain why metals form positive ions.

diamong [38]

Answer:

Explanation:

Metal elements form positively charged ions called cations because they are located on the left side of the periodic table These elements all have valence electrons in an s orbital. These electrons are relatively easy for the atom to lose to achieve a stable octet of electrons in its outermost energy shell.

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