If an atom has 35 protons in the nucleus how many electrons will have orbiting the nucleus

How many atoms of phosphorus are in 8.10 mol of copper(II) phosphate?

madreJ [45]

You know from the chemical formula that 1 mole of Cu3(PO4)2 contains 2 moles of P so you can work out how many moles of P are in 8.1 moles of Cu3(PO4)2.

<span>As for anything, 1 mole of X contains 6.022 * 10^23 of X. Multiply moles of X by Avogadro's number.</span>

7 0

3 years ago

Read 2 more answers

Carmen has a sample of matter. It is clear and smells sour. The sample is also thick but it flows when poured.

Lady bird [3.3K]

Answer:

Explanation:

Answer:

Its a liquid cause it flows

Explanation:

think about it none of the others make sense ice is a solid and its clear

something like a wall can be thick but it doesn't flow

8 0

3 years ago

The average blood volume in the human body is about 4.9 L. Convert this volume to cubic centimeters (cm3).a. milliliters (mL)b.

CaHeK987 [17]

Answer:

a. 4900 mL; b. 4900 cm³ c. 4.9x10⁻³ m³

Explanation:

1 L = 1 dm³

4.9 L = 4.9 dm³

1 dm³ = 1000 cm³ → 4900 cm³

1 cm³ = 1 mL (4900 mL)

1 dm³ = 1x10⁻³ m³ → 4.9x10⁻³ m³

6 0

3 years ago

How much heat is released when 10.0 grams of a substance cools 15 degrees? The specific

VLD [36.1K]

Answer:

Q = 525 J

Explanation:

Given that,

Mass, m = 10 grams

The change in temperature, $\Delta T=15^{\circ} C$

The specific heat of the substance is 3.5 J/g °C

We need to find the amount of heat released in the process. It can be given by the formula as follow :

$Q=mc\Delta T\\\\Q=10\ g\times 3.5\ J/g^{\circ} C\times 15^{\circ} C\\Q=525\ J$

Hence, 525 J of heat is released.

4 0

3 years ago

Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 li

Lera25 [3.4K]

Answer:

4.41

Explanation:

Step 1: Write the balanced equation

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

Step 2: Calculate the respective concentrations

$[CO]_i = \frac{0.500mol}{5.00L} = 0.100M$

$[H_2]_i = \frac{1.500mol}{5.00L} = 0.300M$

$[H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M$

Step 3: Make an ICE chart

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

I 0.100 0.300 0 0

C -x -3x +x +x

E 0.100-x 0.300-3x x x

Step 4: Find the value of x

Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396

Step 5: Find the concentrations at equilibrium

[CO] = 0.100-x = 0.100-0.0396 = 0.060 M

[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M

[CH₄] = x = 0.0396 M

[H₂O] = x = 0.0396 M

Step 6: Calculate the equilibrium constant (Kc)

$Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41$

3 0

4 years ago