Sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. If you have 379.4 grams of calcium chloride and an excess of sodium phosphate, how much calcium phosphate can you make?

Answers

Answer 1

Answer:

Answer:- 353.3 g

Solution:- The balanced equation is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

We start with given grams of calcium chloride and convert them to moles. Then using mol ratio, the moles of calcium phosphate are calculated and converted to grams as.

Molar mass of calcium chloride is 110.98 gram per mol and molar mass of calcium phosphate is 310 gram per mol.

The set is made using dimensional analysis as:

$379.4gCaCl_2((1molCaCl_2)/(110.98gCaCl_2))((1molCa_3(PO_4)_2)/(3molCaCl_2))((310gCa_3(PO_4)_2)/(1molCa_3(PO_4)_2))$

= $353.3gCa_3(PO_4)_2$

So, 353.3 grams of calcium phosphate can be formed.

Answer 2

Answer:

Answer:

353.3g

Explanation:

Related Questions

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Four beakers containing potassium nitrate dissolved in water are allowed to evaporate to dryness. Beakers 1 through 4 contain 2.3, 1.91, 5.985, and 0.52 g of dry potassium nitrate respectively. How many moles of potassium nitrate were recovered after the water evaporated?
Please help me i don’t know the answer to this!!!!!!
The genetic code from DNA is carried into the cytoplasm by____
Which of the following is an example of physical change?

When NaOH is added to water, the (OH) = 0.04 M. What is the [H30*]?What is the PH of the solution?

Answers

Answer:

[H₃O⁺] = 2.5 × 10⁻¹³ M

pH = 12.6

Explanation:

Step 1: Given data

Concentration of OH⁻: 0.04 M

Step 2: Calculate the concentration of H₃O⁺

Let's consider the self-ionization of water reaction.

2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

The ionic product of water is:

Kw = [OH⁻] × [H₃O⁺] = 10⁻¹⁴

[H₃O⁺] = 10⁻¹⁴ / [OH⁻]

[H₃O⁺] = 10⁻¹⁴ / 0.04

[H₃O⁺] = 2.5 × 10⁻¹³ M

Step 3: Calculate the pH

The pH is:

pH = -log [H₃O⁺] = -log 2.5 × 10⁻¹³ = 12.6

3.7500*10^4+9.7100*5

Answers

Answer:

37548.55

Explanation:

3.7500*10^4+9.7100*5

3.7500*10000+9.7100*5

37500+48.55

37548.55

A mixture of neon and xenon gases, at a total pressure of 739 mm Hg, contains 0.919 grams of neon and 19.1 grams of xenon. What is the partial pressure of each gas in the mixture?_______g Xe

Answers

Answer:

Partial pressure of neon = 175 mmHg

Partial pressure of xenon = 564 mmHg

Explanation:

The partial pressure of a gas in a mixture can be calculated as the product of the mole fraction of the gas (Xi) and the total pressure (Pt), as follows:

Pi = Xi Pt

The total pressure is 739 mmHg ⇒ Pt = 739 mmHg

In order to calculate the mole fraction of each gas, we have to first calculate the number of moles of each gas (n) by dividing the mass of the gas into the molar mass (MM):

For neon gas (Ne):

MM(Ne) = 20.18 g/mol

n(Ne)= mass/MM = 0.919 g x 1 mol/20.18 g = 0.045 mol Ne

For xenon gas (Xe):

MM(Xe) = 131.3 g/mol

n(Xe)= mass/MM = 19.1 g x 1 mol/131.3 g = 0.145 mol Xe

Now, we calculate the mole fraction (X) by dividing the number of moles of the gas into the total number of moles (nt):

nt= moles Ne + moles Xe = 0.045 mol + 0.145 mol = 0.190 mol

X(Ne) = moles Ne/nt = 0.045 mol/0.190 mol = 0.237

X(Xe) = moles Xe/nt = 0.145/0.190 mol = 0.763

Finally, we calculate the partial pressures of Ne and Xe as follows:

P(Ne) = X(Ne) x Pt = 0.237 x 739 mmHg = 175 mmHg

P(Xe) = X(Xe) x Pt = 0.763 x 739 mmHg = 564 mmHg

A geochemist in the field takes a 13.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 16.° C, and caps the sample carefully. Back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.143 gRequired:
Using only the information above, can you calculate the solubility of X in water at 15°C ? If you said yes, calculate it.

Answers

Answer:

The answer is YES

The value is $S = 110 \ g/L$

Explanation:

From the question we are told that

The volume of the sample taken is v = 13.0 mL

The temperature is $T = 16^oC$

The mass of the sample is $m = 0.143 g$

Generally the solubility of the substance X is mathematically represented as

$S = (m)/(V)$

=> $S = (0.143 )/(13.0)$

=> $S = 0.011 \ g/mL$

=> $S = 110 \ g/L$

Radioactive gold-198 is used in the diagnosis of liver problems. the half-life of this isotope is 2.7 days. if you begin with a sample of 8.1 mg of the isotope, how much of this sample remains after 2.6 days?

Answers

Answer:

See explanation below

Explanation:

To solve this problem, we need to use the expression of half life decay of concentration (or mass) which is the following:

m = m₀e^-kt (1)

In this case, k will be the constant rate of this element. This is calculated using the following expression:

k = ln2/t₁/₂ (2)

Let's calculate the value of k first:

k = ln2/2.7 = 0.2567 d⁻¹

Now, we can use the expression (1) to calculate the remaining mass:

m = 8.1 * e^(-0.2567 * 2.6)

m = 8.1 * e^(-0.6674)

m = 8.1 * 0.51303

m = 4.16 mg remaining

Final answer:

The half-life of gold-198 is the time it takes for half of it to decay. Given that the half-life is 2.7 days, and the period in consideration is 2.6 days, approximately half of the original amount of 8.1 mg, which is 4.05 mg, will remain.

Explanation:

This problem is related to the concept of half-life in radioactive decay. The half-life of a substance is the time it takes for half of it to decay. As the half-life of gold-198 is 2.7 days and we are considering a period of 2.6 days, which is almost one half-life, therefore, approximately half the substance should have decayed.

So, if you start with 8.1 mg of gold-198, at the end of one half-life (or close to it at 2.6 days), you should have approximately half of this amount remaining. Half of 8.1 mg is 4.05 mg, thus, approximately 4.05 mg remains after 2.6 days.