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mezya [45]
2 years ago
5

What information can a foliated metamorphic rock provide you about the conditions under which it formed?

Chemistry
1 answer:
Mnenie [13.5K]2 years ago
4 0
A foliated rock forms when the minerals are realigned due to the presence of high temperature and pressure. The minerals align in such a way that the axes are directed to where the pressure was applied.
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How many moles of SnF₂ will be produced along with 48 grams of H₂? *
Oxana [17]

Answer:

The correct option is;

24 moles

Explanation:

Here, we have the reaction as follows;

Sn(s) + 2HF(g) → SnF₂ (s) + H₂ (g)

Therefore, one mole of Sn reacts with 2 moles HF to form one mole of SnF₂ and one mole of H₂

Molar mass of H₂ = 2.01588 g/mol

Therefore, the number of moles of H₂ in 48 grams of H₂ is given by the relation;

Number \ of  \ moles \ of \, H_2 = \frac{Mass \ of  H_2}{Molar \ Mass \ of  H_2}=\frac{48}{2.01588} = 23.8109 \ moles \approx 24 \ moles

Since one mole each of SnF₂ and H₂ are produced, the number of moles of SnF₂ produced = 24 moles.

The number of moles of SnF₂ that will be produced is 24 moles.

6 0
2 years ago
How many kilograms of water must be added to 6.07 grams of oxalic acid (H2C2O4) to make a 0.025 m solution?
uranmaximum [27]
This dilution problem uses the equation
M
a
V
a
=
M
b
V
b

M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
4 0
2 years ago
How many electrons in each orbit?
sveticcg [70]
In 1st orbit 2
2nd 8
3rd 10
f orbital has 16
4 0
3 years ago
2 or more forces that are not equal and opposite of one another can cause change in motion
DedPeter [7]

Answer:

D unbalanced force

Explanation:

i think

4 0
2 years ago
The maximum amounts of lead and copper allowed in drinking water are 0.015 mg/kg for lead and 1.3 mg/kg for copper. Tell the max
GuDViN [60]

Answer:

  • <em>The maximum amount of copper allowed in 100 g of water is </em><u><em>0.00013 g</em></u>

Explanation:

To find the maximum amount of copper (in grams) allowed in 100 g of water use the maximum amount ratio (1.3 mg / kg)  and set a proportion with the unknown amount of copper (x) and the amount of water (100 g):

First, convert 100 g of water to kg: 100 g × 1 kg / 1000 g = 0.1 kg.

Now, set the proportion:

  • 1.3 mg Cu / 1 Kg H₂O = x / 0.1 kg H₂O

Solve for x:

  • x = 0.1 kg H₂O × 1.3 mg Cu / 1 kg H₂O = 0.13 mg Cu

Convert mg to grams:

  • 0.13 mg × 1 g / 1,000 mg = 0.00013 g

Answer: 0.00013 g of copper.

6 0
2 years ago
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