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horrorfan [7]
3 years ago
10

A decrease in seawater temperature or an increase in salinity causes

Chemistry
1 answer:
Vinil7 [7]3 years ago
8 0
A decrease in seawater temp
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For the balanced equation shown below, if the reaction of 0.112 grams of
Alekssandra [29.7K]

Answer:

The answer is 74.5%.

Explanation:

As we know that % yield=  \frac{actual yield}{theoretical yield} x 100%.

Therefore,

Step 1 Calculate Theoretical yield:

0.112H_{2} x   \frac{1 mol H_{2} }{2.016 g of H{2} }    x    \frac{4 mol H_{2}O }{4 mol H_{2} }    x  \frac{18.02 g H_{2}O }{1 mol H_{2}O}   = 1.001 g H_{2}O

Now Step 2

% yield   =  \frac{actual yield}{theoretical yield} x 100% =   \frac{0.745g}{1.001g} = 74.5%

8 0
3 years ago
A metal sample weighs 57.3g and has volume of 6.3mL what is the density of the sample
PilotLPTM [1.2K]

Answer:

<h2>9.10 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

density =  \frac{57.3}{6.3}  \\  = 9.095238...

We have the final answer as

<h2>9.10 g/mL</h2>

Hope this helps you

5 0
3 years ago
What is the density of an object with a volume of 12 mL and a mass of 38g?
Andreas93 [3]

Answer:

Hi im an online tutor and I can assist you with all your assignments. we have professionals in all fields. Check out our website https://toplivewriters.com

Explanation:

5 0
2 years ago
The density of air at ordinary atmospheric pressure and 25 ∘C is 1.19 g/L. What is the mass, in kilograms, of the air in a room
marshall27 [118]
Given:
The density of air = 1.19 g/L at 25°C and atmospheric pressure,
or
density = 1.19 x 10⁻³ kg/L

Volume of air in the room is
V = 12.5*19.5*6.0 = 1462.5 ft³

Note that
1 ft³ = 28.317 L
Therefore
V = (1462.5 ft³)*(28.317 L/ft³) = 4.1414 x 10 ⁴ L

By definition, mass = density*volume.
Therefore, the mass is
(1.19 x 10⁻³ kg/L)*(4.1414 x 10⁴ L) = 49.283 kg

Answer: 49.3 kg (nearest tenth)
7 0
3 years ago
How many liters of 0.305 M K3PO4 solution are necessary to completely react with 187 mL of 0.0184 M NiCl2 according to the balan
Mashutka [201]

Answer:

6.55 mL of K₃PO₄ are required

Explanation:

We need to propose the reaction, in order to begin:

2K₃PO₄ (aq) + 3NiCl₂(aq) →  Ni₃(PO₄)₂ (s) ↓ + 6KCl (aq)

Molarity = mol/L (Moles of solute that are contained in 1 L of solution.)

M = mol / volume(L).  Let's find out the moles of chloride:

- We first convert the volume from mL to L → 187 mL . 1L / 1000mL = 0.187L

0.0184 M . 0.187L = 0.00344 moles of NiCl₂

Ratio is 3:2. Let's propose this rule of three:

3 moles of chloride react with 2 moles of phosphate

Then, 0.00344 moles of NiCl₂ will react with (0.00344 . 2) /3 = 0.00229 moles of K₃PO₄

M = mol / volume(L) → Volume(L) = mol/M

Volume(L) = 0.00229 mol / 0.350 M = 6.55×10⁻³L

We convert the volume from L to mL → 6.55×10⁻³L . 1000mL /1L = 6.55 mL

5 0
3 years ago
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