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Kitty [74]
3 years ago
13

Describe two other “medicinal” uses for metals.

Chemistry
1 answer:
dedylja [7]3 years ago
3 0

Answer:

Please see "Explanation" for a description of how and why these two metals work in the medical field.

1. Platinum (Pt) is used in chemotherapy.

2. Nickel (Ni) used in medical tools and equipment.

Explanation:

Two chemotherapy agents contain the heavy metal Platinum. Those agents cause crosslinking in DNA, making it impossible for the dividing cancer cell to duplicate it's DNA, eventually leading to the death of that cancer cell. Platinum allows the DNA's bases to magnetically crosslink, and without platinum, there wouldn't be enough magnetic energy to cause this.

Nickel-containing stainless steel is also widely used in manufacturing medical devices, pharmaceuticals and vaccines, where the highest standards of hygiene and sterility are essential.

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The table compares the final cost of
Snowcat [4.5K]

Answer:

y = -2.5 + x

Explanation:

Given the data:

__X____Y

$20 __$17.50

$21 __ $18.50

$22__ $19.50

$23__ $20.50

Using a linear regression calculator :

The linear regression model obtained is :

y = -2.5 + x

Where ; - 2.5 = intercept ; gradient / slope = 1

3 0
3 years ago
If it requires 48.3 milliliters of 0.55 molar nitric acid to neutralize 15.0 milliliters of barium hydroxide, solve for the mola
Gemiola [76]
First you have to moles so multiply .0483L X .55M= .026565 Multiply moles by mole ratio which is 1/2, so the moles becomes .013283 now molarity=moles/volume; divide .013283/.015L=.885533M significant figures and you final answer is 0.89M
8 0
3 years ago
Read 2 more answers
PLS HELP A gas has a volume of 250 mL at 96.5 kPa. What will the pressure of the gas be if the
ella [17]

Answer:

Solve the following problems (assuming constant temperature). Assume all numbers are 3 sig figs. 1. A sample of oxygen gas occupies a volume of 250 mL at 740 torr pressure. ... the gas exert if the volume was decreased to 2.00 liters? ... A 175 mL sample of neon had its pressure changed from 75.0 kPa to 150 kPa.

Explanation:

8 0
3 years ago
A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following
marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

\frac{V_1}{n_1} =\frac{V_2}{n_2}

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}

<em>V₂ = 4,00L</em>

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}

<em>V₂ = 16,00L</em>

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}

<em>V₂ = 12,31L</em>

I hope it helps!

6 0
4 years ago
Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.
BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>

Now we can find percentage composition / percentage by mass of oxygen.

% composition = \frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound} × 100

% composition = \frac{144}{240.9} × 100 = <u>59.776%</u>

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0
2 years ago
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